Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$ Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?

(I assume that what you mean is that you don't want to use the fact that $\sum 1/n^2=\pi^2/6$.)
Yes, there is a way of seeing this, and this is the basis of Mikael Passare's paper How to compute $\sum 1/n^2$ by solving triangles
(free preprint on arXiv,
or the published version on JSTOR for subscribers).
If you set $x=e^{-t}$, your integral becomes
$$
\int_0^{\infty} -\ln(1-e^{-t}) \, dt
,
$$
or
$$
\int_0^{\infty} -\ln(1-e^{-x}) \, dx
$$
if we call the variable $x$ again.
This is the area of a region in the first quadrant of the $xy$-plane, below the curve $y=-\ln(1-e^{-x})$ (or equivalently $e^{-x}+e^{-y}=1$). This is the region called $U_0$ i Passare's paper, and he shows that there are two other regions $U_1$ and $U_2$ with the same area, and then via a clever area-preserving change of variables that the combined area of $U_0$, $U_1$ and $U_2$ equals the area of a certain right triangle $T$ which is obviously $\pi^2/2$. Hence the area of $U_0$ (your integral) is $\pi^2/6$.
(See also this answer and this.)
