Derive$~2(\frac{a}{a^2-x^2}+\frac{b}{b^2-x^2})\left(\frac{(a+x)(b+x)}{(a-x)(b-x)}\right)=\frac{(a+b+2x)(a-x)(b-x)+(a+x)(b+x)(a+b-2x)}{(a-x)^2(b-x)^2}$

$$y=\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\tag{1}$$

$$=\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}$$

$$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\right)\tag{2}$$

$$u:=\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}~~\leftarrow~~\text{Preparation for chain rule}\tag{3}$$

$$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt[4]{u}\right)$$

$$=\frac{d}{du}\left(\sqrt[4]{u}\right)\frac{du}{dx}$$

$$=\frac{d}{du}\left(u^{\frac{1}{4}}\right)\frac{du}{dx}$$

$$=\frac{1}{4}u^{-\frac{3}{4}}\frac{du}{dx}$$

$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)$$

$$f\left(x\right):=\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)$$

$$g\left(x\right):=\left(a+x\right)\left(b+x\right)$$

$$h\left(x\right):=\left(a-x\right)\left(b-x\right)$$

$$\therefore~~f\left(x\right)=\frac{g'\left(x\right)h\left(x\right)-g\left(x\right)h'\left(x\right)}{h^{2}\left(x\right)}$$

$$g'\left(x\right)=\left(b+x\right)+\left(a+x\right)=a+b+2x$$

$$h'\left(x\right)=\left(-1\right)\left(b-x\right)+\left(-1\right)\left(a-x\right)$$

$$=\left(-1\right)\left(\left(b-x\right)+\left(a-x\right)\right)$$

$$=-\left(a+b-2x\right)$$

$$f\left(x\right)=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)-\left(a+x\right)\left(b+x\right)\left(-1\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}$$

$$=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}=\frac{du}{dx}$$

$$\frac{dy}{dx}=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)}_{=\frac{du}{dx}}$$

$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}}_{=:\text{green}\left(x\right)}$$

$$=\underbrace{\frac{1}{2}\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)}_\text{What I want to finally derive. }$$

$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)$$

$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{4}{4}}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{4}{4}}$$

$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{4}{4}}$$

$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{}\frac{2}{2}$$

$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{2}{1}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{}}_{\text{I want to derive this from}~\text{green}\left(x\right)}$$

$$\text{green}\left(x\right)=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}$$

$$=\frac{1}{\left(a-x\right)^{2}\left(b-x\right)^{2}}\left\{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\frac{\left(a+x\right)}{\left(a+x\right)}\left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)}{\left(a^{2}-x^{2}\right)\left(b-x\right)}\left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)}{\left(a^{2}-x^{2}\right)\left(b-x\right)} \frac{\left(b-x\right)^{-1}}{\left(b-x\right)^{-1}} \left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)\left(b-x\right)^{-1}}{\left(a^{2}-x^{2}\right)} \left(a+b-2x\right)\right\}$$

$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{}\left(b+x\right)^{}\left(a+x\right)\left(b-x\right)^{-1}}{\left(a^{2}-x^{2}\right)} \left(a+b-2x\right)\right\}$$

Stucked.

BTW the derivation of name of green function is that I had painted the eqn and text with green in markdown however somehow MSE editor can't compile it correctly so I just removed that paining code .

You could have taken $\log$ on both sides to differentiate it efficiently.

$\begin{align}&\log y =\frac14(\log(a+x)+\log(b+x)-\log(a-x)-\log(b-x))\\\\\Rightarrow&\frac1yy' = \frac14\left(\frac1{a+x}+\frac1{b+x}+\frac1{a-x}+\frac1{b-x}\right) =\frac14\left(\frac{2a}{a^2-x^2}+\frac{2b}{b^2-x^2}\right) \\=&\frac12\left(\frac{a}{a^2-x^2}+\frac{b}{b^2-x^2}\right)\\\\\Rightarrow&y' = y\cdot\frac12\left(\frac{a}{a^2-x^2}+\frac{b}{b^2-x^2}\right) = \frac12\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\left(\frac{a}{a^2-x^2}+\frac{b}{b^2-x^2}\right)\end{align}$

as you needed.

I think I have a much shorter proof that only needs algebraic manipulation, let me know if there's something you cannot understand.

$$LHS \\= 2\left(\frac{a}{a^2-x^2}+\frac{b}{b^2-x^2}\right)\left(\frac{(a+x)(b+x)}{(a-x)(b-x)}\right)\\ = \left(\frac{1}{a-x}+\frac{1}{b+x}+\frac{1}{b-x}+\frac{1}{a+x}\right)\left(\frac{(a+x)(b+x)}{(a-x)(b-x)}\right)\\ = \color{red}{\frac{(a+x)(b+x)}{(a-x)^2(b-x)}}+\color{blue}{\frac{(a+x)}{(a-x)(b-x)}}+\color{red}{\frac{(a+x)(b+x)}{(a-x)(b-x)^2}}+\color{blue}{\frac{(b+x)}{(a-x)(b-x)}}\\ = \color{red}{\frac{(a+x)(b+x)(b-x)+(a+x)(b+x)(a-x)}{(a-x)^2(b-x)^2}}+\color{blue}{\frac{(a+x)+(b+x)}{(a-x)(b-x)}}\\ = \frac{\color{blue}{(a+b+2x)(a-x)(b-x)}+\color{red}{(a+x)(b+x)(a+b-2x)}}{(a-x)^2(b-x)^2}\\ = RHS$$