A sketch (this approach is generalizable to arbitrary even powers of the logarithm):
We use the standardbranch of $\log$ throughout
$$
\Re \log^2(1-e^{2 ix})=\log^2(2 \sin(x))-(x-\frac{\pi}{2})^2 \quad (\star)
$$
Integrating the LHS over a large rectangle in the upper half plane with verticies $\{(0,0),(\frac{\pi}{6},0),(\frac{\pi}{6},i\infty),(0,i\infty)\}$ we obtain (the integral over the imaginary line cancels, since we are only interested in real components)
$$
\Re\int_0^{\pi/6}\log^2(1-e^{2 ix})=-\Im\int_0^{\infty}\log^2(1-ae^{-2x})=-\Im(G(a))
$$
where $a=e^{-i \pi/3}$. Substituting $e^{-2x}=q$ we get
$$
2 G(a)=\int_0^1 \frac{\log^2(1-a q)}{q}
$$
using repeated integration by parts we get ($\text{Li}_n(z)$ denotes the Polylogarithm of order $n$)
$$
2 G(a)=-2\text{Li}_3(1-a)+2\text{Li}_2(1-a)\log(1-a)+\log(1-a)^2\log(a)+2\text{Li}_3(1)
$$
adding some Polylogarithm wizardy we find that (see Appendix)
$$
\Im(G(a))=-\color{blue}{\frac{\pi^3}{324}}
$$
using furthermore the trivial integral
$$
\int_0^{\pi/6}dx(x-\frac{\pi}{2})^2=\color{green}{\frac{19 \pi^3}{648}}
$$
we find indeed
$$
\int_0^{\pi/6}dx\log^2(2 \sin(x))=\\
\color{green}{\frac{19 \pi^3}{648}}+\color{blue}{\frac{\pi^3}{324}}=\frac{7\pi^3}{216}
$$
which is equivalent to the claim in question
Following OP we might rewrite the integral as an infinte sum, which gives us the hardly to believe corollary
$$
\sum_{n\geq 0}\frac{1}{16^n(2n+1)^3}\binom{2n}{n}=\frac{7\pi^3}{216}
$$
Appendix
The functional equations of the Dilogarithm immediately delievers
$$\Re\text{Li}_2(1-a)= \frac{1}{2}(\text{Li}_2(1-a)+\text{Li}_2(1-a^{-1})\\=-\frac{1}{4}\log^2(a)=\frac{\pi^2}{36}$$
The Trilogarithm part is a bit trickier,
$\Im\text{Li}_3(1-z)=\Im\int_0^{1-z}\text{Li}_2(x)/x=-\Im\text{Li}_3(z)$ together with the inversion formula
$\text{Li}_3(-z)-\text{Li}_3(-1/z)=-\frac{1}{6}(\log^3(z)+\pi^2\log(z))$gives us that
$$
\Im\text{Li}_3(1-a)=\frac{5\pi^3}{162}
$$
since $\log(1-a)=\frac{i\pi}{3}$ we find
$$
2\Im(G(a))=-2\frac{5\pi^3}{162}+2\frac{\pi^2}{36}\frac{\pi}{3}+\frac{\pi^2}9\frac{\pi}3
$$
or
$$
\Im(G(a))=-\frac{\pi^3}{324}
$$
Best Answer
Applying a substitutiong of $\,\large\frac{1-x}x=t \Rightarrow dx=-\frac{dt}{(1+t)^2}$ gives: $$\int_\frac12^1\frac{\ln x}{x\cdot\frac{1-x}{x}}dx=\int_0^1 \frac{\ln\left({\frac{1}{1+t}}\right)}{t(1+t)}dt$$ $$=\int_0^1 \frac{\ln(1+t)}{1+t}dt-\int_0^1 \frac{\ln(1+t)}{t}dt$$ $$=\frac{\ln^2(1+t)}{2}\bigg|_0^1 - \frac{\pi^2}{12}=\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$$ The second integral can be found here.