Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation . 
Help appreciated.
I've made some research in MSE and found several questions which could be helpful: 
1) $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=-\frac{17\pi^4}{1440}$
2) $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{\pi^6}{12960}$ 
3) $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}=  \frac{1}{3}\zeta(6)$
 A: 
The double summation is equal to
  $$\frac{11\zeta(4)}{8}=\frac{11\pi^4}{720}.$$

Note that
$$\frac{m^2+n^2}{m n\left(m^2-n^2\right)^2}=
\frac{1}{2 mn(m+n)^2}+
\frac{1}{2 mn(m-n)^2}.$$
Now consider the Tornheim double sums:
$$T(a,b,c)= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} 
\frac{1}{m^an^b(m+n)^c}.$$
Then
$$ \sum_{m,n=1\, m\neq n}^{\infty} \frac{1}{mn(m+n)^2}=T(1,1,2)-\sum_{m=1}^{\infty}\frac{1}{4m^4}=T(1,1,2)-\frac{\zeta(4)}{4},$$
$$\begin{align} \sum_{m=1}^{\infty} \sum_{n=1}^{m-1} 
\frac{1}{mn(m-n)^2}
&= \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} 
\frac{1}{mn(m-n)^2}\\
&= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} 
\frac{1}{nk^2(n+k)}=T(1,2,1)\end{align}$$
and again
$$ \sum_{m=1}^{\infty} \sum_{n=m+1}^{\infty} 
\frac{1}{mn(m-n)^2}=T(1,2,1).$$
Hence
$$\begin{align}\sum_{m,n=1\, m\neq n}^{\infty}{\frac{m^2+n^2}{mn(m^2-n^2)^2}}
&=\frac{T(1,1,2)-\zeta(4)/4}{2} +T(1,2,1)\\
&=\frac{11\zeta(4)}{8}
\end{align}$$
where we used 
$$T(1,1,2)=\zeta(4)/2\quad,\quad T(1,2,1)=5\zeta(4)/2$$
(see page 31 in The evaluation of Tornheim double sums).
A: Partial Fractions gives
$$
\frac{m^2+n^2}{\left(m^2-n^2\right)^2}=\frac{1/2}{(m-n)^2}+\frac{1/2}{(m+n)^2}\tag1
$$
First, we will compute
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m-n)^2}
&=2\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{mn(m-n)^2}\tag{2a}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{(m+n)nm^2}\tag{2b}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{m^3}\tag{2c}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{nm^2}\tag{2d}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac{m+n}{nm^3}\tag{2e}\\
&=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{m^2n^2}\tag{2f}\\[3pt]
&=\frac{\pi^4}{36}\tag{2g}
\end{align}
$$
Explanation:
$\text{(2a)}$: symmetry between $m\lt n$ and $n\lt m$
$\text{(2b)}$: substitute $m\mapsto m+n$
$\text{(2c)}$: partial fractions
$\text{(2d)}$: swap $m$ and $n$ in $\text{(2b)}$ then partial fractions
$\text{(2e)}$: average $\text{(2c)}$ and $\text{(2d)}$
$\text{(2f)}$: simplify
$\text{(2g)}$: evaluate $\zeta(2)^2$
Next, we will compute
$$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2}
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\color{#C00}{\frac1{nm^2(n+m)}}-\color{#090}{\frac1{m^2(n+m)^2}}\right)\tag{3a}\\
&=\color{#C00}{\frac{\pi^4}{72}}-\color{#090}{\frac{\pi^4}{120}}\tag{3b}\\[3pt]
&=\frac{\pi^4}{180}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: partial fractions
$\text{(3b)}$: the red sum is $\frac12$ of $\text{(2b)}$, the green sum is $\frac12\left(\zeta(2)^2-\zeta(4)\right)$
$\text{(3c)}$: simplify
Therefore,
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m+n)^2}
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2}-\frac14\zeta(4)\tag{4a}\\
&=\frac{\pi^4}{180}-\frac{\pi^4}{360}\tag{4b}\\[9pt]
&=\frac{\pi^4}{360}\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: subtract the terms where $m=n$
$\text{(4b)}$: apply $\text{(3c)}$
$\text{(4c)}$: simplify
Thus.
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^2+n^2}{mn\left(m^2-n^2\right)^2}
&=\frac12\frac{\pi^4}{36}+\frac12\frac{\pi^4}{360}\tag{5a}\\
&=\frac{11\pi^4}{720}\tag{5b}
\end{align}
$$
Explanation:
$\text{(5a)}$: apply $(1)$, $(2)$, and $(4)$
$\text{(5b)}$: simplify
