Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$ 
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$

My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx
\end{align*}
\begin{align*}
I_x(y)&=\int_0^1\frac{-x\ln(xy)}{2(1-xy)}-\frac{x\ln(xy)}{2(1+xy)} \, dx\\[6pt]
&=\frac{1}{2}\int_0^1\frac{-x\ln(xy)}{1-xy}\, dx+\frac{1}{2} \int_0^1\frac{-x\ln(xy)}{1+xy} \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n \int_0^1x^{n+1}\ln(xy)\,dx+\frac{1}{2} \sum_{n=0}^\infty(-1)^{n+1} \int_0^1y^nx^{n+1}\ln(xy) \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n\left(\frac{\ln(y)}{n+2}-\int_0^1 \frac{x^{n+1}}{(n+2)y} \, dx\right)+\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+1} y^n \left(\frac{\ln(y)}{n+2}-\int_0^1\frac{x^{n+1}}{(n+2)y} \, dx\right)\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty \frac{y^n\ln(y)}{n+2} ((-1)^{n+1}-1) +\frac{1}{2} \sum_{n=0}^\infty \frac{y^{n-1}}{(n+2)^2}(1-(-1)^{n+1})
\end{align*}
 A: 
I thought it might be instructive to present an approach that uses elementary calculus tools to reduce the double integral to a single integral.
Then, after a standard use of a Taylor series expansion of $\log(1+y)$, the answer is expressed as the familiar alternating series of reciprocal squares $\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$.
To this end, we now proceed.

Let $I$ be the double integral of interest given by
$$I=\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dy\,dx$$
Enforcing the substitution $y\mapsto y/x$ in the inner integral reveals
$$\begin{align}
I&\overbrace{=}^{y\mapsto y/x}-\int_0^1 \int_0^x \frac{\log(y)}{1-y^2}\,dy\,dx\tag1
\end{align}$$

We change the order or integration in $(1)$ to find that
$$\begin{align}
I&=-\int_0^1 \int_0^x \frac{\log(y)}{1-y^2}\,dy\,dx\\\
&=-\int_0^1 \int_y^1 \frac{\log(y)}{1-y^2}\,dx\,dy\\\\
&=-\int_0^1 \frac{\log(y)}{1+y}\,dy\tag2
\end{align}$$

It is interesting to note that the integral on the right-hand side of $(2)$ can be recognized as $-\text{Li}_2(-1)=\frac{\pi^2}{12}$ and we are done!


Integrating by parts the integral on the right-hand side of $(2)$, we find that
$$\begin{align}I&=-\int_0^1 \frac{\log(y)}{1+y}\,dy\\\\
&\overbrace{=}^{\text{IBP}}\int_0^1 \frac{\log(1+y)}{y}\,dy\tag3
\end{align}$$

Using the Taylor series for $\log(1+y)=\sum_{n=1}^\infty \frac{(-1)^{n-1}y^n}{n}$ in $(3)$ yields
$$\begin{align}
I&=\int_0^1 \frac{\log(1+y)}{y}\,dy\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\\\\
&=\frac{\pi^2}{12}
\end{align}$$
as was to be shown!
A: \begin{align}
-\int_0^1\int_0^1\frac{x\ln{(xy)}}{1-x^2y^2}\,\mathrm{d}x\,\mathrm{d}y
&=-\sum_{n=0}^\infty \int_0^1 y^{2n} \int_0^1x^{2n+1} \ln{(xy)} \, \mathrm{d}x \, \mathrm{d}y \\[6pt]
&=-\sum_{n=0}^\infty\int_0^1y^{2n}\left[\frac{x^{2(n+1)}\ln{(xy)}}{2(n+1)}-\frac{x^{2(n+1)}}{4(n+1)^2}\right]_0^1\,\mathrm{d}y\\[6pt]
&=-\sum_{n=0}^\infty\int_0^1y^{2n}\left(\frac{\ln{(y)}}{2(n+1)}-\frac1{4(n+1)^2}\right)\,\mathrm{d}y\\[6pt]
&=-\sum_{n=0}^\infty\left[\frac{y^{2n+1}\ln{(y)}}{2(n+1)(2n+1)}-\frac{(4n+3)y^{2n+1}}{4(n+1)^2(2n+1)^2}\right]_0^1\\[6pt]
&=\sum_{n=0}^\infty\frac{4n+3}{4(n+1)^2(2n+1)^2}\\[6pt]
&=\sum_{n=0}^\infty\left(\frac1{(2n+1)^2}-\frac1{(2(n+1))^2}\right)\\[6pt]
&=\sum_{n=1}^\infty\left(\frac1{(2n-1)^2}-\frac1{(2n)^2}\right)\\[6pt]
&=\eta(2)\\[6pt]
&=\frac{\pi^2}{12}
\end{align}
A: \begin{align}
\int _0^1\int _0^1-\frac{x\ln \left(xy\right)}{1-x^2y^2}\:dx\:dy
&=\int _0^1\int _0^1\left(\underbrace{-\frac{x\ln \left(x\right)}{1-x^2y^2}-\frac{x\ln \left(y\right)}{1-x^2y^2}}_{t=x^2}\right)\:dx\:dy
\\[3mm]
&=\int _0^1\int _0^1\left(-\frac{1}{4}\underbrace{\frac{\ln \left(t\right)}{1-ty^2}}_{K}-\frac{1}{2}\frac{\ln \left(y\right)}{1-ty^2}\right)\:dt\:dy
\\[3mm]
&=\frac{1}{4}\int _0^1\frac{\:\operatorname{Li}_{2}\left(y^2\right)}{y^2}\:dy+\frac{1}{2}\int _0^1\frac{\ln \left(y\right)\ln \left(1-y^2\right)}{y^2}\:dy
\\[3mm]
&=\frac{1}{4}\sum _{k=1}^{\infty }\frac{1}{k^2\left(2k-1\right)}+\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k\left(2k-1\right)^2}
\\[3mm]
&=\ln \left(2\right)-\frac{1}{4}\zeta \left(2\right)+\frac{3}{4}\zeta \left(2\right)-\ln \left(2\right)
\\[3mm]
&=\frac{1}{2}\zeta \left(2\right)
\end{align}
I used $\displaystyle \int _0^1\frac{a\ln ^n\left(t\right)}{1-at}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(a\right)$ to evaluate $K$
