Finding the $2n$ th derivative of $\frac{1}{1+x^2y^2}$ with repect to $x$ We know that with repect to $x$,
\begin{align}
\left( \frac{1}{1-x^2 y^2} \right)^{(2n)} = 
\frac{(2n)!y^{2n}}{2 } \left(  \frac{1}{(1+xy)^{2n+1}} +\frac{1}{(1-xy)^{2n+1}} \right)
\end{align}
What about
\begin{align}
\left( \frac{1}{1+x^2 y^2} \right)^{(2n)} 
\end{align}

The first thing to do is make this substitution $y = ik$, we can do this freely because the derivative is with respect to x not to y, and it is only a contsant.Then after this substitution we get
$$\left( \frac{1}{1+x^2 y^2} \right)^{(2n)}=\left( \frac{1}{1-x^2 k^2} \right)^{(2n)}
$$
Which, as you said, is equal to
$$\frac{(2n)!k^{2n}}{2 } \left(  \frac{1}{(1+xk)^{2n+1}} +\frac{1}{(1-xk)^{2n+1}} \right)
$$
Finally we just substitute $\frac{y}{i}=k$ and we obtain that 
for the derivative of a number with the form 4k we have
$$\frac{(2n)!y^{2n}}{2 } \left(  \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right)
$$
and for the derivative of a number with the form 4k+2 we have
$$\frac{-(2n)!y^{2n}}{2 } \left(  \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right)
$$
then  if we do not want i, we  only have to  add the two terms and apply the binomial theorem.
