My solution is a little extensive, but something different here. Let's start by noting that $\displaystyle 2\Re\left\{\ln(1-ix)\right\}=\ln\left(1+x^2\right)$, so we can write:
\begin{equation*}
\int_0^1\frac{\ln(1+x)\ln\left(1+x^2\right)}{x}\mathrm dx=\Re\left\{\int_0^1\frac{2\ln(1+x)\ln(1-ix)}{x}\mathrm dx\right\},\tag{1}
\end{equation*}
using $2ab=a^2+b^2-(a-b)^2$ with $a=\ln(1+x)$ and $b=\ln(1-ix)$, the inner integral brackets in (1) is:
\begin{align*}
\int_0^1\frac{2\ln(1+x)\ln(1-ix)}{x}\mathrm dx&=\int_0^1\left\{\ln^2(1-ix)+\ln^2(1+x)-\ln^2\left(\frac{1+x}{1-ix}\right)\right\}\frac{\mathrm dx}{x}\\
&\hspace{-3cm}=\int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx+\int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx-\int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}.\tag{2}
\end{align*}
Let's evaluate $\displaystyle \int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx$, so let $1-ix\mapsto x$:
\begin{align*}
\int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx&=\int_{1-i}^1\frac{\ln^2(x)}{1-x}\mathrm dx\\
&=\sum_{n=1}^\infty\int_{1-i}^1 x^{n-1}\ln^2(x)\mathrm dx\\
&=\sum_{n=1}^\infty\frac{\mathrm d^2}{\mathrm dn^2}\left[\frac1n-\frac{(1-i)^n}{n}\right]\\
&\hspace{-2cm}=2\zeta(3)-\sum_{n=1}^\infty\left[\ln^2(1-i)\frac{(1-i)^n}{n}-2\ln(1-i)\frac{(1-i)^n}{n^2}+2\frac{(1-i)^n}{n^3}\right]\\
&=2\zeta(3)+\ln^2(1-i)\ln(i)+2\ln(1-i)\text{Li}_2(1-i)-2\text{Li}_3(1-i),\tag{3}
\end{align*}
where we have the values:
\begin{align*}
\begin{cases}
\ln^2(1-i)\ln(i)&=\displaystyle\frac{3}{4}\zeta(2)\ln(2)-\left(\frac{\pi^3}{32}-\frac\pi 8\ln^2(2)\right)i\\
\\
2\ln(1-i)\text{Li}_2(1-i)&=\displaystyle-\frac{\pi G}{2}-\frac{3}{8}\zeta(2)\ln(2)-2\left(\frac12 G\ln(2)+\frac{\pi^3}{64}+\frac\pi 8\ln^2(2)\right)i\\
\\
2\Re\left\{\zeta(3)-\text{Li}_3(1-i)\right\}&=\displaystyle\frac{29}{32}\zeta(3)-\frac38\zeta(2)\ln(2)
\end{cases}
\end{align*}
The above values can be derived using the Trilog and Dilog identities:
\begin{align*}
\therefore\quad\text{Li}_3(z)+\text{Li}_3(1-z)+\text{Li}_3\left(1-\frac1z\right)&=\zeta(3)+\frac{\ln^3(z)}{6}+\zeta(2)\ln(z)-\frac{\ln^2(z)\ln(1-z)}{2}\\
\text{Li}_2(z)+\text{Li}_2(1-z)&=\zeta(2)-\ln(z)\ln(1-z)
\end{align*}
and the $\displaystyle\ln(i)=\frac{i\pi}{2}$, $\displaystyle\ln(1-i)=\frac12\ln(2)-\frac{i\pi}{4}$ values. As we are dealing with the real part of the values above then $\Re\left\{\text{Li}_3(1-i)+\text{Li}_3(\overline{1-i})\right\}=2\Re\left\{\text{Li}_3(1-i)\right\}$, leaving the third equation of the system justified. Therefore:
\begin{equation*}
\therefore\ \Re\left\{\int_0^1\frac{\ln^2(1-ix)}{x}\mathrm dx\right\}=-\frac{\pi G}2+\frac{29}{32}\zeta(3).\tag{4}
\end{equation*}
Now, for $\displaystyle \int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx$ we have straightforward the result:
\begin{equation*}
\therefore\quad\int_0^1\frac{\ln^2(1+x)}{x}\mathrm dx=\frac14\zeta(3),\tag{5}
\end{equation*}
in which it can be easily calculated.
For $\displaystyle \int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}$ in (2), take the replacement $\displaystyle \frac{1+x}{1-ix}\mapsto x$ yielding:
\begin{align*}
\int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}&=(1+i)\int_{1+i}^1\frac{(1+ix)\ln^2(x)}{(1-x)(1+ix)^2}\mathrm dx\\
&=(1+i)\int_{1+i}^1\frac{\ln^2(x)}{1+i}\left(\frac{1}{1-x}+\frac{i}{1+ix}\right)\mathrm dx\\
&=\int_{1+i}^1\frac{\ln^2(x)}{1-x}\mathrm dx+i\int_{1+i}^1\frac{\ln^2(x)}{1+ix}\mathrm dx\\
&=\sum_{n=1}^\infty \frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx-\sum_{n=1}^\infty(-i)^n\frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx,\tag{6}
\end{align*}
where, the terms in common in both sums above, are:
\begin{equation*}
\therefore\quad\frac{\mathrm d^2}{\mathrm dn^2}\int_{1+i}^1 x^{n-1}\mathrm dx=-\frac{\ln^2\left(1+i\right)(1+i)^n}{ n}-\frac{2(1+i)^n}{n^3}+\frac{2}{n^3}+\frac{2\ln\left(1+i\right)(1+i)^n}{n^2}.
\end{equation*}
Applying the summation in (6), we have:
\begin{align*}
&=\ln^2(1+i)\ln(-i)-2\text{Li}_3\left(1+i\right)+2\ln(1+i)\text{Li}_2(1+i)+2\zeta(3)\\
&\quad-\ln^2(1+i)\ln\left(i\right)+2\text{Li}_3\left(1-i\right)-2\ln(1+i)\text{Li}_2\left(1-i\right)-2\text{Li}_3(-i)\\
&=\ln^2(1+i)\left\{\ln(-i)-\ln(i)\right\}+2\ln(1+i)\left\{\text{Li}_2(1+i)-\text{Li}_2(1-i)\right\}\\
&\quad+2\left\{\text{Li}_3(1-i)-\text{Li}_3(1+i)\right\}+2\zeta(3)-2\text{Li}_3(-i)
\end{align*}
Assuming the following values:
\begin{align*}
\begin{cases}
\ln^2(1+i)\left\{\ln(-i)-\ln(i)\right\}&=\displaystyle \frac32\zeta(2)\ln(2)+\frac1{16}\left(\pi^3-4\pi\ln^2(2)\right)i\\
\\
\displaystyle 2\ln(1+i)\left\{\text{Li}_2(1+i)-\text{Li}_2(1-i)\right\}&=\displaystyle-\pi G-\frac32\zeta(2)\ln(2)+\left(2G\ln(2)+\frac\pi 2\ln^2(2)\right)i\\
\\
\displaystyle 2\Re\left\{\text{Li}_3(1-i)-\text{Li}_3(1+i)\right\}&=2\Re\left\{\text{Li}_3(1-i)-\text{Li}_3(\overline{1-i})\right\}=0\\
\\
\displaystyle 2\left\{\zeta(3)-\text{Li}_3(-i)\right\}&=\displaystyle \frac{35}{16}\zeta(3)-\frac{\pi^3}{16}i.
\end{cases}
\end{align*}
We have finally:
\begin{equation*}
\therefore\ \Re\left\{\int_0^1\ln^2\left(\frac{1+x}{1-ix}\right)\frac{\mathrm dx}{x}\right\}=-\pi G+\frac{35}{16}\zeta(3).\tag{7}
\end{equation*}
Collecting (7), (5) and (4) so taking (2) and consequently (1), the results follows naturally:
\begin{align*}
\therefore\quad\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}\mathrm dx=\frac{\pi G}{2}-\frac{33}{32}\zeta(3).\qquad\blacksquare
\end{align*}
