A tricky summation

Similar solution with @Guilherme Niedu 's.

$\begin{aligned} S & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {m^2n}{3^m(3^mn+3^nm)} & \small \color{#3D99F6} \text{Add another }S \text{ interchanging }m \text{ and }n. \\ & = \frac 12 \left(\sum_{m=1}^\infty \sum_{n=1}^\infty \frac {m^2n}{3^m(3^mn+3^nm)} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {mn^2}{3^n(3^nm+3^mn)} \right) \\ & = \frac 12 \left(\sum_{m=1}^\infty \sum_{n=1}^\infty \frac {3^nm^2n+3^m mn^2}{3^{m+n}(3^mn+3^nm)} \right) \\ & = \frac 12 \left(\sum_{m=1}^\infty \sum_{n=1}^\infty \frac {mn\left(3^nm+3^mn\right)}{3^{m+n}(3^mn+3^nm)} \right) \\ & = \frac 12 \left(\sum_{m=1}^\infty \sum_{n=1}^\infty \frac {mn}{3^{m+n}} \right) \\ & = \frac 12 \left(\sum_{m=1}^\infty \frac m{3^m} \sum_{n=1}^\infty \frac n{3^n} \right) & \small \color{#3D99F6} \text{See note: }\sum_{n=1}^\infty \frac n{3^n} = \frac 34 \\ & = \frac 12 \times \frac 34 \times \frac 34 = \frac 9{32} = \boxed{0.28125} \end{aligned}$

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Note:

$\begin{aligned} S_n & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{3^n} = \sum_{\color{#D61F06}n=0}^\infty \frac n{3^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{3^{n+1}} = \frac 13 {\color{#3D99F6}\sum_{n=0}^\infty \frac n{3^n}} + \frac 13 \sum_{n=0}^\infty \frac 1{3^n} = \frac 13 {\color{#3D99F6}S_n} + \frac 12 = \frac 34 \end{aligned}$

First of all, let us consider the sum:

$\large \displaystyle \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}, |x| < 1$

Differentiating with respect to $x$ and then multiplying by $x$ on both sides:

$\color{#20A900} \boxed{\large \displaystyle \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x) ^2}, |x| < 1}$

Now, to the main sum:

$\large \displaystyle S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n} {3^m (n3^m + m3^n)}$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n} {3^m (n3^m + m3^n)} + \frac{n^2 m} {3^n (m3^n + n3^m)} \right]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{3^n m^2 n + 3^m n^2 m} {3^{m+n} (n3^m + m3^n)} \right]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn(n3^m + m3^n) } {3^{m+n} (n3^m + m3^n)} \right]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn} {3^{m+n}} \right]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \frac{m}{3^{m}} \sum_{n=1}^{\infty} \frac{n} {3^{n}} \right]$

Using the above result (highlighted in green) in each sum, with $x = \frac13$ :

$\large \displaystyle S = \frac12 \left [ \frac34 \cdot \frac34 \right]$

$\color{#3D99F6} \boxed{\large \displaystyle S = \frac{9}{32} = 0.28125}$