How can I prove this double sum equation?

Question

I'm trying to prove that: $$\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{(m^2+n^2x^2)^2}=-\frac{\pi}{2x}-\frac{1}{x^2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{\left(m^2+n^2/x^2\right)^2}$$ I'm missing $-\frac{\pi}{2x}$ in my process, just by swapping $n\to m,\, m\to n$ in sum above on LHS as follows: $$\begin{align*}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2x^2-m^2}{(m^2+n^2x^2)^2}&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2x^2-n^2}{(n^2+m^2x^2)^2}\\ &=-\frac1{x^2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{n^2/x^2-m^2}{(m^2+n^2~/x^2)^2}\end{align*}$$ Here I see that the exchange of summation operators probably cannot be applied, but I can't see that. Any help will be welcome!