What is the proof of the following: ∫01(ln⁡t1−t)2dt=π23 ?

This is by no means a complete solution but a possible route.
Letting ${\displaystyle t=\frac{1}{x}}$ note that
${\displaystyle I={\int }_0^1{\left(\frac{\ln t}{1-t}\right)}^{2}dt}$
${\displaystyle ={\int }_1^{\mathrm{\infty }}{\left(\frac{\ln t}{1-t}\right)}^{2}dt}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}{\left(\frac{\ln (1+t)}{t}\right)}^{2}dt}$
Setting ${\displaystyle 1+t=e^x}$, we get
${\displaystyle I={\int }_0^{\mathrm{\infty }}\frac{x^2}{{(e^x-1)}^2}{e}^{x}dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{x^2}{{(e^{\frac{x}{2}}-e^{-\frac{x}{2}})}^2}dx}$
${\displaystyle =2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2}{{\text{sinh}}^2\left(x\right)}dx}$
The last integral can be done by the method of residues to get
${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2}{{\text{sinh}}^2\left(x\right)}dx=\frac{{\pi }^2}{6}}$
I will fill this in once I get back home.

${\displaystyle {\int }_0^1{\left(\frac{\log t}{1-t}\right)}^2={\int }_0^{\mathrm{\infty }}\frac{x^2{e}^x}{{(e^x-1)}^2}dx}$
Heres

Consider
$\sum _{k=0}^{\mathrm{\infty }}{t}^k=\frac{1}{1-t}$ (1)
Differentiating (1) with respect to t yields
$\sum _{k=1}^{\mathrm{\infty }}k{t}^{k-1}=\frac{1}{(1-t{)}^2}$ (2)
Multiplying both sides of (2) by ${\ln }^{2}t$ and integrating from t=0 to t=1 yields
${\int }_0^1(\frac{\ln t}{1-t}{)}^{2}dt={\int }_0^1\sum _{k=1}^{\mathrm{\infty }}k{t}^{k-1}{\ln }^{2}tdt$
$=\sum _{k=1}^{\mathrm{\infty }}k{\int }_0^1{t}^{k-1}{\ln }^{2}tdt$ (3)
Using formula
${\int }_0^1{x}^{\alpha }{\ln }^{n}xdx=\frac{(-1{)}^{n}n!}{(\alpha +1{)}^{n+1}},\text{ for }n=0,1,2,...$ (4)
then (3) becomes
${\int }_0^1(\frac{\ln t}{1-t}{)}^{2}dt=2\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$
$=\frac{{\pi }^2}{3}$
where $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\zeta (2)=\frac{{\pi }^2}{6}$