Some Sum

$\sum_{x=1}^{\infty }\sum_{y=1}^{\infty }\frac{1}{xy(x+y)}= \sum_{x=1}^{\infty }\sum_{y=1}^{\infty }\frac{1}{xy} \int_0^\infty e^{-(x+y)t} \text{dt}= \int_0^\infty \left( \sum_{x=1}^{\infty } \dfrac{e^{-xt}}{x} \right)\left(\sum_{y=1}^\infty \dfrac{e^{-yt}}{y} \right) \text{dt} =\int_0^\infty \ln^2 \left( 1-e^{-t}\right) \text{dt}$ by making a change of variable with from $t \to 1-e^{-t}$ , the integral is $= \int_0^1 \dfrac{\ln^2 (t)}{1-t} \text{dt} = \sum_{n=0}^\infty \int_0^1 t^n \ln^2(t) \text{dt}=\sum_{n=0}^\infty \dfrac{\partial^2}{\partial n^2}\int_0^1 t^n \text{dt} =\sum_{n=0}^\infty \dfrac{2}{(n+1)^3} = \boxed{2\zeta(3)}$