**Faster method to calculate the exact solution of the following integral (based on the ideas of Fedor Petrov: https://mathoverflow.net/users/4312/fedor-petrov, Double integral with logarithms, URL (version: 2019-04-15): https://mathoverflow.net/q/328126):**

$$J\equiv \int_{0}^{1}{\int_{0}^{1}{\frac{\ln x-\ln y}{x-y}}}dxdy .$$

Since $$f\left( x,y \right)=\frac{\ln x-\ln y}{x-y}=f\left( y,x \right),$$ the surface $f\left( x,y \right) $ is symmetric with respect to the bisector plane $x = y$; so,

$$\frac{J}{2}=\int_{0}^{1}{dx\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}}dy.$$

With the change of variable$$y\equiv tx,\ t\in \left( 0,\ 1 \right),$$ the integral $$\int_{0}^{x}{\frac{\ln x-\ln y}{x-y}}dy,$$ is transformed into the following one that does not depend on $x$, $$ I\equiv -\int_{0}^{1}{\frac{\ln t}{1-t}\,}dt.$$

The integration on the unit square $(0, 0), (1, 0), (1, 1), (0,1)$ is reduced to the integration on the triangle $(0, 0), (1, 0), (1, 1).$

To solve the integral $I$ we will carry out the new change of variable,
$$s\equiv 1-t,$$
by which $I$ is transformed into the integral that defines the *dilogarithm*, whose value for $s = 1$ coincides with the *Riemann zeta* $\zeta \left( 2 \right)$,whose value is well known:

$$I=\int_{1}^{0}{\frac{\ln \left( 1-s \right)}{s}\,}ds=\text{L}{{\text{i}}_{2}}\left( 1 \right)=\zeta \left( 2 \right)=\frac{{{\pi }^{2}}}{6}.$$ Therefore, the solution to the proposed integral is $$J=\frac{{{\pi }^{2}}}{3}.$$

**Note**. I've tried it by *polylogarithmic* transformations, but I couldn't get the result $\frac{{{\pi }^{2}}}{3}.$