Let’s start with $$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$
We are going to handle the integral by first integrating $\arctan x.$
$\displaystyle \begin{aligned}\int \operatorname{arctanh} x dx & =x \operatorname{arctanh} x-\int \frac{x}{1-x^2} \\& =x \operatorname{arctanh} x-\frac{1}{2} \ln \left(1-x^2\right) \\& =\frac{1}{2}\left[x \ln \left(\frac{1+x}{1-x}\right)-\ln \left(1-x^2\right)\right]\end{aligned}\tag*{} $
Then performing integration by parts yields
$\displaystyle \begin{aligned}&2 \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \\= & \int_0^1 \arctan x d\left(x \ln \left(\frac{1+x}{1-x}\right)-\ln \left(1-x^2\right)\right) \\= & \underbrace{ \int_0^1 \frac{x}{1+x^2} \ln \left(\frac{1-x}{1+x}\right) d x}_{J} - \underbrace{ \int_0^1 \frac{\ln \left(1-x^2\right)}{1+x^2} d x}_{K} \end{aligned}$
For both integrals $J$ and $K$, we use the substitution $t=\frac{1-x}{1+x}$, then
$\displaystyle \begin{aligned}& J=\int_0^1 \frac{\frac{1-t}{1+t}}{\frac{2\left(1+t^2\right)}{(1+t)^2}} \ln t \cdot \frac{2 d t}{(1+t)^2} \\& =\int_0^1 \frac{(1-t) \ln t}{(1+t)\left(1+t^2\right)} d t \\& =\int_0^1 \frac{\ln t}{t+1} d t-\int_0^1 \frac{t \ln t}{t^2+1} d t \\& =\int_0^1 \frac{\ln t}{t+1} d t-\frac{1}{4} \int_0^1 \frac{\ln t}{t+1} d t \\& =\frac{3}{4} \int_0^1 \frac{\ln t}{t+1} d t \\& =\frac{3}{4}\left(-\frac{\pi^2}{12}\right) \\& =-\frac{\pi^2}{16} \\\end{aligned}\tag*{} $
Similarly for $K$, we have
$\displaystyle \begin{aligned}K& =\int_0^1 \frac{\ln \frac{4 t}{(1+t)^2}}{\frac{2\left(1+t^2\right)}{(1+t)^2}} \cdot \frac{2 d t}{(1+t)^2} \\& =\int_0^1 \frac{\ln (4 t)-2 \ln (1+t)}{1+t^2} d t \\& =\ln 4 \int_0^1 \frac{d t}{1+t^2}+\int_0^1 \frac{\ln t}{1+t^2} d t-2 \int_0^1 \frac{\ln (1+t)}{1+t^2}dt \\& =\frac{\pi \ln 4}{4}-G-\frac{\pi}{4} \ln 2 (\textrm{ For details, please refer to } \cdots (*))\\& =\frac{\pi}{4} \ln 2-G\end{aligned}\tag*{} $
Now we can conclude that
$\displaystyle \boxed{\int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y =\frac{\pi^2}{32}+\frac{\pi}{8} \ln 2-\frac{G}{2}}\tag*{} $
Footnote (*)
We first express the integrand in terms of $\tan \theta$.
$\displaystyle \int_0^1 \frac{\ln (1+t)}{1+t^2} d t=\int_0^{\frac{\pi}{4}} \ln (\tan \theta+1) d \theta\tag*{} $
Then using the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, we have
$$\displaystyle \begin{aligned}\int_0^1 \frac{\ln (1+t)}{1+t^2} d t& =\int_0^{\frac{\pi}{4}} \ln \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta \\& =\int_0^{\frac{\pi}{4}} \ln \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta \\& =\int_0^{\frac{\pi}{4}} \ln 2 d \theta-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\\int_0^1 \frac{\ln (1+t)}{1+t^2} d t & =\frac{\pi \ln 2}{8 }\end{aligned}\tag*{} $$