Streak

To make things easier, we can rewrite $f(x)$ as

$f(x)=\dfrac{1}{1-x^2}= \sum_{k=0}^{\infty}x^{2k}$

With radius of convergence $|x|<1$

Since we have to evaluate the $n$ -th derivative at $x=0$ , we can use this latter formula

$f(x)=1+x^2+x^4+x^6+\cdots$

Since $n$ is even, there is a term of the form $x^n$

If we derive $n$ times, al terms from $1$ up to $x^{n-2}$ wanish. We now proceed to find the $n$ -th derivative of $x^n$

$\dfrac{d(x^n)}{dx}=nx^{n-1}$

$\dfrac{d^2(x^n)}{dx^2}=n(n-1)x^{n-2}$

$\vdots$

$\dfrac{d^n(x^n)}{dx^n}=n(n-1)(n-2)\cdots 2\cdot 1=n!$

Since all the terms from $x^{n+2}$ onwards after the procedure have at least a factor of $x$ , when we plug in $0$ they vanish too, making the answer

$f^{(n)}(0)=\boxed{n!}$