$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m²n}{n3^m +m3^n}$ $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m²n}{n3^m +m3^n}$.

I replaced m by n,n by m and sum both which gives term $\frac{mn(m+n)}{n3^m +m3^n}$.how to do further?


A closely related summation doable by hand is when $3^m$ multiplies the denominator of OP:
$$S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m²n}{3^m(n3^m +m3^n)}~~~(1)$$ $$S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{a_m(a_m+a_n)},~~ a_k=3^k/k~~~(2)$$
Interchange $m$ and $n$, then
$$S=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{a_n(a_n+a_m)}~~~(3)$$
Adding (2) and (3), we get
$$2S=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{a_m a_n} =\left(\sum_ {k=0}^{\infty} \frac{1}{a_k}\right)^2= \left(\sum_ {k=0}^{\infty} \frac{k}{3^k}\right)^2.$$
Next use $$\sum_{k=1}^{\infty} kx^k=\frac{x}{(1-x)^2}, {x}<1.$$
$$\implies S=\frac{1}{2} \frac{9}{16}=\frac{9}{32}$$
