Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$.

You may write $z = xy\Rightarrow z'(x) = y$.

$$f(x) = \frac{y^{2n+1}xy}{1-x^2y^2} = y^{2n+1}\color{blue}{\frac{z}{1-z^2}} $$ $$\Rightarrow f(x) = g(z(x)) \Rightarrow f'(x) = g'(z)\cdot z'(x) = g'(z)\cdot y \Rightarrow \boxed{f^{(k)}(x) = g^{(k)}(z)\cdot y^k}$$

$$\color{blue}{\frac{z}{1-z^2} = -\frac{1}{2}\left( \frac{1}{z+1}+ \frac{1}{z-1} \right)}$$

$$\Rightarrow f^{(2n+1)}(x) = g^{(2n+1)}(z)\cdot y^{2n+1} = -\frac{y^{2(2n+1)}}{2}\left( \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z+1)^{2n+2}}+ \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z-1)^{2n+2}} \right) $$ $$= \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(z+1)^{2n+2}}+ \frac{1}{(z-1)^{2n+2}} \right) = \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(xy+1)^{2n+2}}+ \frac{1}{(xy-1)^{2n+2}} \right)$$