Does the above function have a closed form?
If $$f(x)=\sum_{n=1}^\infty \frac{x^2}{\sinh^2(xn)} \, ,$$ then we can find its value at $x=0$ as follows: For any $\epsilon>0$ and $N\in\mathbb{N}$ pick $0<x<\epsilon/N$ s.t. $$f(x)=\sum_{n=1}^N \frac{x^2}{\sinh^2(xn)} + \sum_{n=N+1}^\infty \frac{x^2}{\sinh^2(xn)} =\sum_{n=1}^N \frac{1}{n^2} \frac{1}{1+O(x^2n^2)} + \sum_{n=N+1}^\infty \frac{x^2}{\sinh^2(xn)} \\ =\sum_{n=1}^N \frac{1}{n^2} + \underbrace{O(x^2N)}_{=O(\epsilon^2/N)} + \underbrace{\sum_{n=N+1}^\infty \frac{x^2}{\sinh^2(xn)}}_{\leq\int_N^\infty \frac{x^2}{\sinh^2(xn)} \, {\rm d}n \\\stackrel{t=xn}{=} \int_{xN}^\infty \frac{x}{\sinh^2(t)} \, {\rm d}t \\ =O(x/(xN))=O(1/N)} = \sum_{n=1}^N \frac{1}{n^2} + O(1/N) \, .$$ Since $N$ was arbitrary, the result $$\lim_{x\rightarrow 0} f(x) = \sum_{n=1}^\infty \frac{1}{n^2} = \zeta(2)$$ follows.
However, in the complex vicinity of $x=0$, $f(x)$ has a rather preculiar behaviour, namely that there are poles of second order at $x=\frac{i\pi m}{n}$, where $m=\pm 1,\pm 2,...$ and $n\in \mathbb{N}$, so there are poles at rational multiples of $\pi$ on the imaginary number line.
It appears, that $f(x)$ does not have a laurent expansion about $x=0$. Calculating the coefficients $c_n$ of $f(x)=\sum\limits_{n\in\mathbb{Z}} c_n x^n$, I basically get $c_n=0$ for almost all $n$. This is reasonable, as the big essential wall singularity on the imaginary axis seems to segregate the regions $\Re(x)>0$ and $\Re(x)<0$. Nevertheless, concentrating on $\Re(x)>0$, it seems (numerically) that $f(x)=\zeta(2)-x+O(x^2)$ for small real $x>0$.
Furthermore it is easy to show $f'(x)<0$ for all $x>0$ and $\lim\limits_{x\rightarrow \infty} f(x)=0$.
