This is a personal research that came to an end , since the results were not those which were being anticipated. I was unable to come up with a solution therefore I post the topic here:
Prove (it certainly holds because it was checked on a computer) that the following identity holds:
$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^2}=\zeta(2)\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\left ( 2n-1 \right )^2}-\zeta(4)$$
wheras $\zeta$ represents the zeta function defined as $\displaystyle \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}, \;\; \mathfrak{Re}(s)>1 $. Of course both values of $\zeta$ appearing here are known. For complete of sakeness i quote them :
$$\zeta(2)=\frac{\pi^2}{6}, \;\; \zeta(4)=\frac{\pi^4}{90}$$
Now, one can also see (not trivial ) that:
$$\sum_{n=1}^{\infty}\frac{1}{n^2 \sinh^2 \pi n}=\frac{4}{\pi^2}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^2} -\frac{\pi^2}{60}$$
This equation also holds (checked by computer). The following result was extracted by using the known formulae $\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{z+n}=\frac{\pi}{\tan \pi z}$ and the known (?) Fourier series: $$\displaystyle \frac{1}{\sinh^2 \pi z}=\frac{1}{\pi^2 z}+\frac{4z^2}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{\left ( z^2+k^2 \right )^2}-\frac{2}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{z^2+k^2} $$
Of course the last sum at the last equation can easily be computed via residues. What remains now is the proof for the first equation in the post. No-one can guarantee that is going to be an easy task.
Some comments:
1. I came across the identity on a book. I checked the validity with a computer and yes it does hold.
2. I have ckecked enough books to see if there is in there somewhere , but unfortunately it was not. So, I speculate that it is not that famous.
3. It can also be linked with other sums (single or double). Unfortunately I don't have my papers in front of me in order to write them down. So, i think it is an interseting identity.
I would appreciate your help.