A difficult double integral $\int_{0}^{1}\int_{0}^{1}\frac{x\ln x \ln y }{1-xy}\frac{dxdy}{\ln(xy)}$

How to evaluate $$\int_{0}^{1}\int_{0}^{1}\frac{x\ln x\ln y}{1-xy}\frac{dxdy}{\ln(xy)} ?$$

Any ideas on how to even start with this integral? It seems impossible to me.

There's a similar integral that originates from this site.

Here's an approach that should work for similar integrals too, such as this one.

Mostly the idea is to use Feynman's trick in two dimensions. Consider the following integral:

$$I(n)=\int_0^1\int_0^1 \frac{(xy)^{n-1} x\ln x\ln y}{\ln(xy)}\mathrm dx\,\mathrm dy.$$Differentiating with respect to $n$ gives $$I'(n)=\int_0^1\int_0^1 (xy)^{n-1} x \ln x \ln y \,\mathrm dx\,\mathrm dy\qquad\qquad\qquad\qquad\quad$$ $$=\int_0^1 x^{n} \ln x \,\mathrm dx \int_0^1 y^{n-1} \ln y\,\mathrm dy=\frac{1}{(n+1)^2}\frac{1}{n^2}.$$ Now we have to to get back to $I(n)$. Since $I(\infty)=0$, we have that $$I(n)=-(I(\infty)-I(n))=-\int_n^\infty \frac{1}{(x+1)^2 x^2 }\mathrm dx=-\frac{1}{n}-\frac{1}{n+1}+2\ln\left(1+\frac{1}{n}\right).$$ Finally, notice that $$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\int_0^1\int_0^1 \frac{ x\ln x\ln y}{(1-xy)\ln(xy)}\,\mathrm dx\,\mathrm dy=\sum_{n=1}^\infty \int_0^1\int_0^1 \frac{(xy)^{n-1} x\ln x\ln y}{\ln(xy)}\mathrm dx\,\mathrm dy$$ $$ =-\sum_{n=1}^\infty \left(\underbrace{\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)}_{=\gamma}+\underbrace{\frac{1}{n+1}-\frac{1}{n}}_{=-1}+\underbrace{\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)}_{=\gamma}\right)=1-2\gamma.$$ See here for the above.