By symmetry we have $J/2=\int_0^1 dx \int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-\int_0^1 \frac{\log t} {1-t}dt$. It does depend on $x$ and is well known to be equal to $\pi^2/6$ (you may use the geometric progression expansion $\frac{1}{1 - t} =\sum_{n>0 } t^{n-1}$ and integrate term-wise to get $\sum 1/n^2$).

By symmetry we have $J/2=\int_0^1 dx \int_0^x f(x, y) dy$ where $f(x, y) $ is your integrand. Integrating against $y$ for fixed $x$ we denote $y=tx$, $t$ varies from 0 to 1 and the integral against $y$ reads as $-\int_0^1 \frac{\log t} {1-t}dt$. It does not depend on $x$ and is well known to be equal to $\pi^2/6$ (you may use the geometric progression expansion $\frac{1}{1 - t} =\sum_{n>0 } t^{n-1}$ and integrate term-wise to get $\sum 1/n^2$).