Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$. $f(x,y) = \frac{y^{2n+1}xy}{1-x^2y^2}$.
I made the following table:
\begin{align}
& 2n+1 = 1 \implies f^{(1)} = \frac{1!y^2(1+x^2y^2)}{(1-x^2y^2)^2}\\
& 2n+1 = 3 \implies f^{(3)} = \frac{3!y^6(x^4y^4+6x^2y^2+1)}{(1-x^2y^2)^4} \\
& 2n+1 = 5 \implies f^{(5)} = \frac{5!y^{10}(x^6y^6+15x^4y^4+15x^2y^2+1)}{(1-x^2y^2)^6} \\
& 2n+1 = 7 \implies f^{(7)} = \frac{7!y^{14}(x^8y^8+28x^6y^6+70x^4y^4+28x^2y^2+1)}{(1-x^2y^2)^8}
\end{align}
I think $f^{(2n+1)}$ has the shape:
\begin{align}
f^{(2n+1)} = \frac{(2n+1)!y^{2(2n+1)}P(x,y)}{(1-x^2y^2)^{2n+2}}
\end{align}
The problem is finding $P(x,y)$ above.

You may write $z = xy\Rightarrow z'(x) = y$. 
$$f(x) = \frac{y^{2n+1}xy}{1-x^2y^2} = y^{2n+1}\color{blue}{\frac{z}{1-z^2}} $$
$$\Rightarrow f(x) = g(z(x)) \Rightarrow f'(x) = g'(z)\cdot z'(x) = g'(z)\cdot y \Rightarrow \boxed{f^{(k)}(x) = g^{(k)}(z)\cdot y^k}$$
$$\color{blue}{\frac{z}{1-z^2} = -\frac{1}{2}\left( \frac{1}{z+1}+ \frac{1}{z-1} \right)}$$
$$\Rightarrow f^{(2n+1)}(x) = g^{(2n+1)}(z)\cdot y^{2n+1} = -\frac{y^{2(2n+1)}}{2}\left( \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z+1)^{2n+2}}+ \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z-1)^{2n+2}} \right) $$ $$=  \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(z+1)^{2n+2}}+ \frac{1}{(z-1)^{2n+2}} \right) = \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(xy+1)^{2n+2}}+ \frac{1}{(xy-1)^{2n+2}} \right)$$
