How to evaluate $\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$ As seen in the title I'm interested in a way to evaluate
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$
But I'm not sure what to do, I did attempt something but ended up with a wrong result
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k-\sqrt{2}n\right)\left(k+\sqrt{2}n\right)}$$
then applying partial fraction decomposition
\begin{align*}
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k-\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3\left(k+\sqrt{2}n\right)}+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^2n^2}\\
&=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k-\sqrt{2}n-1}\:\mathrm{d}x+\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{k^3}\int _0^1x^{k+\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{-\sqrt{2}n-1}\:\mathrm{d}x-\sum _{n=1}^{\infty }\int _0^1\operatorname{Li}_3\left(-x\right)x^{\sqrt{2}n-1}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x-\int _0^1\frac{x^{\sqrt{2}}\operatorname{Li}_3\left(-x\right)}{x\left(1-x^{\sqrt{2}}\right)}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=\int _0^1\frac{\operatorname{Li}_3\left(-x\right)}{x}\:\mathrm{d}x+\frac{5}{4}\zeta (4)\\
&=-\frac{7}{8}\zeta (4)+\frac{5}{4}\zeta (4)\\
&=\frac{3}{8}\zeta (4)
\end{align*}
Which seems to be a wrong answer due to numerical methods and approximations, what did I do wrong? how else could I approach this problem?

Let $S_1(n)=\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{ k^2-2n^2} $, then $S=\sum\limits_{n=1}^{\infty }\left (\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{n^2\left ( k^2-2n^2 \right )} \right )=\sum\limits_{n=1}^{\infty }\frac{S_1(n)}{n^2}$
Let's consider the integral in the complex plane along a big circle  of radius $R$ (counter-clockwise) with the center at $(0;0)$:
$\oint=\oint_C\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}dz=2\pi i \sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}$ inside the circle
It can be shown that $\oint\to0$ as $R\to\infty\,\,\, \Rightarrow$
$\sum Res\frac{\pi}{\sin\pi z}\frac{1}{z^2-2n^2}=\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2} +\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}=0$
$-\sum\limits_{k=-\infty}^{\infty } \frac{(-1)^k}{k^2-2n^2}=2S_1(n)+\frac{1}{2n^2}=\frac{\pi}{\sqrt2 n\sin(\sqrt2 \pi n)}\,\,\,\Rightarrow\,\,\,S_1(n)=\frac{\pi}{2\sqrt2 n\sin(\sqrt2 \pi n)}-\frac{1}{4n^2}\,\,\,\Rightarrow$
$$S=\sum\limits_{n=1}^{\infty }\frac{\pi}{2\sqrt2 n^3\sin(\sqrt2 \pi n)}-\frac{1}{4}\zeta(4)$$
The evaluation of $\sum\limits_{n=1}^{\infty }\frac{1}{n^3\sin(\sqrt2 \pi n)}=-\frac{13 \pi^3}{360 \sqrt{2}}$ is done here: Compute the series $\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$
NB  The approach is the same: integration along the closed contour in the complex plane with the special integrand $\biggl(\frac{1}{z^3\sin(\pi z)\sin(\sqrt2\pi z-\pi z)}\biggr)$
Finally,
$$S=-\frac{13\pi^4}{360*4}-\frac{\pi^4}{360}=-\frac{17\pi^4}{1440}$$
