\begin{align*} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac1{mn^2+m^2n+2mn} &= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac1{mn(m+n+2)} \\ &= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\int_0^1\frac{x^{m+n+1}}{mn}\;dx \\ &= \int_0^1\left[\sum_{m=1}^{\infty}\frac{x^m}m\right]\left[\sum_{n=1}^{\infty}\frac{x^n}n\right]x\;dx \\ &= \int_0^1x\log^2(1-x)\;dx \\ &= -\int_0^1\frac12(1-x^2)\frac{2\log(1-x)}{1-x}\;dx \\ &= -\int_0^1(1+x)\log(1-x)\;dx \\ &= \int_0^1\frac{\frac32-x-\frac12x^2}{1-x}\;dx \\ &= \int_0^1\left(\frac32 + \frac12x\right)\;dx \\ &= \frac32 + \frac14 \\ &= \frac74 \end{align*}So since all the terms are positive, any partial sum will be less than $\frac74$. $\blacksquare$

I have to post this although it's easy. \begin{align*} \sum_{m=1}^N \sum_{n=1}^N \frac{1}{mn^2+m^2n+2mn} &= \sum_{m=1}^N \left( \frac{1}{m(m+2)} \sum_{n=1}^N \left( \frac{1}{n} - \frac{1}{n+m+2} \right) \right) \\ &< \sum_{m=1}^N \left( \frac{1}{m(m+2)} \left(1 + \frac{1}{2} + \cdots + \frac{1}{m+2} \right) \right) \\ &= \frac{1}{2} \left( \sum_{m=1}^N \frac{1}{m} \left( 1 + \frac{1}{2} + \cdots + \frac{1}{m+2} \right) \right. \\ &\quad \left. - \sum_{m=1}^N \frac{1}{m+2} \left( 1 + \frac{1}{2} + \cdots + \frac{1}{m+2} \right) \right) \\ &= \frac{1}{2} \left[ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) \right] \\ &\quad + \frac{1}{2} \sum_{m=3}^N \left[ \frac{1}{m} \left( 1 + \cdots + \frac{1}{m+2} \right) - \frac{1}{m} \left( 1 + \cdots + \frac{1}{m} \right) \right] \\ &= \frac{23}{16} + \frac{1}{2} \sum_{m=3}^N \frac{1}{m} \left( \frac{1}{m+1} + \frac{1}{m+2} \right) \\ &= \frac{23}{16} + \frac{1}{2} \sum_{m=3}^N \left[ \left( \frac{1}{m} - \frac{1}{m+1} \right) + \frac{1}{2} \left( \frac{1}{m} - \frac{1}{m+2} \right) \right] \\ &< \frac{23}{16} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{4} \right) \right) \\ &= \frac{7}{4}. \end{align*}

\begin{align*} \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty}\dfrac{1}{mn(m+n+2)} &=\sum\limits_{m=1}^{\infty}\left(\frac{1}{m(m+2)}\sum\limits_{n=1}^{\infty}(\dfrac{1}{n}-\frac{1}{m+n+2})\right)\\ &=\sum\limits_{m=1}^{\infty}\frac{1}{m(m+2)}(1+\frac12+\cdots+\frac{1}{m+2})\\ &=(1+\frac12)\sum\limits_{m=1}^{\infty}\frac{1}{m(m+2)}+\sum\limits_{k=3}^{\infty}(\frac{1}{k}\sum\limits_{j=k}^{\infty}\frac{1}{(j-2)j})\\ &=\frac{3}{2}\cdot\frac{3}{4}+\sum\limits_{k=3}^{\infty}\frac{1}{2k}(\frac{1}{k-2}+\frac{1}{k-1})\\ &=\frac{9}{8}+\frac{1}{2}(\frac{3}{4}+\frac{1}{2})\\ &=\frac{7}{4} \end{align*}

What about $$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^2 n + m n^2 + rmn}$$

Bump this