Show that $\int_0^1 \int_0^1 {x\ln x\over (1-xy)\ln(xy)} \, dx \, dy=1-\gamma.$ 
$$\int_{0}^{1}\int_{0}^{1}{x\ln x\over (1-xy)\ln(xy)} \, dx \, dy=1-\gamma.\tag1$$

Let $u=xy$
$$\int_0^1 {1\over y^2}\int_0^y {u\ln u -u\ln y \over (1-u)\ln(u)}dudy\tag2$$
$$\int_0^y {u\over 1-u} \, du-\ln y\int_0^y {u\over \ln u} \, du\tag3$$
$$-y-\ln(1-y)-\ln y \int_0^y {u\over \ln u} \, du\tag4$$
As for this integral
Setting $n=1$
$$f(n,u)=\int_0^y {u^n\over \ln u} \, du\tag5$$
We can remove $\ln u$ by differentiating 
$${df\over dn}=\int_0^y u^n \, du = {y^{n+1}\over n+1}\tag6$$
How can I move on to the next step?
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\begin{align}
&\color{#f00}{%
\int_{0}^{1}\int_{0}^{1}{x\ln\pars{x} \over \pars{1 - xy}\ln\pars{xy}}
\,\dd x\,\dd y} =
\int_{0}^{1}\int_{0}^{y}{\pars{x/y}\ln\pars{x/y} \over
\pars{1 - x}\ln\pars{x}}\,{\dd x \over y}\,\dd y
\\[3mm] = &\
\int_{0}^{1}{1 \over \pars{1 - x}\ln\pars{x}}
\int_{x}^{1}\bracks{x\ln\pars{x}\,{1 \over y^{2}} - x\,{\ln\pars{y} \over y^{2}}}\,\dd y\,\dd x
\\[3mm] = &\
\int_{0}^{1}{1 \over \pars{1 - x}\ln\pars{x}}\braces{%
x\ln\pars{x}\pars{-1 + {1 \over x}} - x\bracks{1 - x + \ln\pars{x} \over x}}\,\dd x
\\[3mm] = &\
\int_{0}^{1}\bracks{1 - {1 - x + \ln\pars{x} \over \pars{1 - x}\ln\pars{x}}}
\,\dd x =
1\ -\ \underbrace{\int_{0}^{1}{1 - x + \ln\pars{x} \over \pars{1 - x}\ln\pars{x}}}
_{\ds{\color{#f00}{\large ?} = \color{#f00}{\large\gamma}}} 
\,\dd x =
\color{#f00}{1 - \gamma}
\end{align}

\begin{align}
\color{#f00}{\large ?} & =
\int_{0}^{1}{1 - x + \ln\pars{x} \over \pars{1 - x}\ln\pars{x}}\,\dd x
=
-\int_{0}^{1}{\pars{x - 1}/\ln\pars{x} - 1 \over 1 - x}\,\dd x =
-\int_{0}^{1}{1 \over 1 - x}\int_{0}^{1}\pars{x^{t} - 1}\,\dd t\,\dd x
\\[3mm] & =
\int_{0}^{1}\int_{0}^{1}{1 - x^{t} \over 1 - x}\,\dd x\,\dd t =
\int_{0}^{1}\bracks{\Psi\pars{t + 1} + \gamma}\,\dd t =
\ln\pars{\Gamma\pars{2}} - \ln\pars{\Gamma\pars{1}} + \gamma =
\color{#f00}{\gamma}
\end{align}


Note that
  $$
\int_{0}^{1}\pars{x^{t} - 1}\,\dd t =
{1 \over \ln\pars{x}}\int_{0}^{1}x^{t}\ln\pars{x}\,\dd t - 1 =
{1 \over \ln\pars{x}}\int_{0}^{1}\partiald{x^{t}}{t}\,\dd t - 1 =
{x - 1 \over \ln\pars{x}} - 1
$$

A: You may prove first that:
$$\iint_{(0,1)^2}\frac{x^{k+1}y^k \log(x)}{\log(xy)}\,dx\,dy = -\frac{1}{k+2}+\log(k+2)-\log(k+1)\tag{1}$$
through differentiation under the integral sign or other techniques, then by summing the RHS of $(1)$ over $k\geq 0$ we get:
$$ I = \iint_{(0,1)^2}\frac{x\log x}{(1-xy)\log(xy)}\,dx\,dy = 1-\sum_{k\geq 1}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)=1-\gamma\tag{2}$$
as wanted. 
