Which rule or method can be applied against$~\lim_{x\to0}\frac{\sqrt{1+6x}-\left(1+3x\right)}{x^{2}}~?$

$$L:=\lim_{x\to0}\frac{\sqrt{1+6x}-\left(1+3x\right)}{x^{2}}~~\leftarrow~~\frac{0}{0}~\text{form} \tag{1} $$

$$=\lim_{x\to0}\frac{\left(1+6x\right)^{\frac{1}{2}}-1-3x}{x^{2}}\tag{2}$$

$$=\lim_{x\to0}\frac{\frac{1}{2}\frac{6}{\sqrt{1+6x}}-3}{2x}$$

$$=\lim_{x\to0}\frac{\frac{3}{\sqrt{1+6x}}-3}{\frac{2x}{1}}$$

$$=\lim_{x\to0}\frac{1}{2x}\left(\frac{3}{\sqrt{1+6x}}-3\right)$$

$$=\lim_{x\to0}\left(\frac{3}{2x\sqrt{1+6x}}-\frac{3}{2x}\right)$$

$$=\lim_{x\to0}\frac{3}{2x\sqrt{1+6x}}-\frac{3}{2}\lim_{x\to0}\frac{1}{x}$$

$$=\frac{3}{2}\lim_{x\to0}\frac{1}{x\sqrt{1+6x}}-\frac{3}{2}\lim_{x\to0}\frac{1}{x}$$

$$=\frac{3}{2}\left(\lim_{x\to0}\frac{1}{x\sqrt{1+6x}}-\lim_{x\to0}\frac{1}{x}\right)$$

$$=\frac{3}{2}\left(\underbrace{\lim_{x\to0}\frac{1}{x}}_{\text{Divergence}}\underbrace{\lim_{x\to0}\frac{1}{\sqrt{1+6x}}}_{=1}-\underbrace{\lim_{x\to0}\frac{1}{x}}_{\text{Divergence}}\right)$$

Stucked. I need your help.

$$L=\lim_{x\to0}\frac{\sqrt{1+6x}-\left(1+3x\right)}{x^{2}}$$

$$=\lim_{x\to0}\frac{\left(\sqrt{1+6x}-\left(1+3x\right)\right)}{x^{2}}\frac{\left(\sqrt{1+6x}+\left(1+3x\right)\right)}{\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$$

$$=\lim_{x\to0}\frac{\left(1+6x\right)-\left(1+3x\right)^{2}}{x^{2}\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$$

$$=\lim_{x\to0}\frac{1+6x-\left(1+6x+9x^{2}\right)}{x^{2}\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$$

$$=\lim_{x\to0}\frac{1+6x-1-6x-9x^{2}}{x^{2}\left(1+3x+\sqrt{1+6x}\right)}$$

$$=\lim_{x\to0}\frac{-9x^{2}}{x^{2}\left(1+3x+\sqrt{1+6x}\right)}$$

$$=\lim_{x\to0}\frac{-9}{\left(1+3x+\sqrt{1+6x}\right)}$$

$$=\lim_{x\to0}\frac{-9}{\left(1+1\right)}=-\frac{9}{2}$$

From your line 4:

$$\lim_{x\to0}\frac{\frac{3}{\sqrt{1+6x}}-3}{2x} \to \dfrac{0}{0}$$

So by L'Hopital's Rule,

$$\lim_{x\to0}\frac{\frac{3}{\sqrt{1+6x}}-3}{2x} = \lim_{x\to0}\frac{-\frac{3}{2(1+6x)^{3/2}}\cdot 6}{2} $$

$$= \dfrac{\frac{-3}{2\cdot(1+6(0))^{3/2}}\cdot 6}{2}=-\dfrac{9}{2}$$

To make life easier when considering $$A=\frac{\sqrt{1+6x}-\left(1+3x\right)}{x^{2}}$$ let $$\sqrt{1+6x}=y \implies x=\frac{1}{6} \left(y^2-1\right)$$

Replace and simplify to make $$A=-\frac{18}{(y+1)^2}$$ and now, just compute the value when $y=1$.