Is possible to use "Feynman's trick" (differentiate under the integral or Leibniz integral rule) to calculate $\int_0^1 \frac{\ln(1-x)}{x}dx\:?$

A direct application might be

$$\left.\frac{d}{ds} \int_0^{1-\delta} \frac{x^s}{1-x} \, dx\right|_{s = 0} = \left.\int_0^{1-\delta} \frac{ x^s \ln x}{1-x} \, dx\right|_{s = 0} = \int_0^{1-\delta} \frac{\ln x}{1-x} \, dx = \int_\delta^{1 } \frac{\ln (1-x)}{x} \, dx $$

Evaluate the integral on the LHS using the geometric series expansion expansion of $1/(1-x)$ and then take the limit as $\delta \to 0$ (since the improper integral on the RHS converges).

You can also proceed by proving the second equality in

$$- \int_0^1 \frac{\ln(1-x)}{x} \,dx = \int_0^1 \int_0^1 \frac{1}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{k^2} =\zeta(2) = \frac{\pi^2}{6}$$

using the geometric series $1/(1 -xy) = 1 + xy + (xy)^2 + \ldots $ The first equality is fairly obvious.

More generally we get by the same process

$$\int_0^1 \int_0^1 \frac{x^\alpha y^\alpha}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{(k + \alpha)^2}$$

and Feynman's trick of repeated integration with respect to $\alpha$ is used to extend the result to other integrals.

Let $\displaystyle J=\int_0^1 \frac{\ln(1-x)}{x}\,dx$

Let $f$ be a function defined on $\left[0;1\right]$,

$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\cos t-s}{\sin t}\right)\,dt$

Observe that,

$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t}{\sin t}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2}-t\right)\,dt\\ &=\left[\frac{t(\pi-t)}{2}\right]_0^{\frac{\pi}{2}}\\ &=\frac{\pi^2}{8} \end{align}$

$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\cos t-1}{\sin t}\right)\,dt\\ &=\int_0^{\frac{\pi}{2}}\arctan\left(-\tan\left(\frac{t}{2}\right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\ &=-\frac{\pi^2}{16} \end{align}$

For $0<s<1$,

$\begin{align} f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1-2s\cos t+s^2}\,dt\\ &=-\Big[\frac{\ln(1-2s\cos t+s^2)}{2s}\Big]_0^{\frac{\pi}{2}}\\ &=\frac{\ln(\left(1-s)^2\right)}{2s}-\frac{\ln(1+s^2)}{2s}\\ &=\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\\ \end{align}$

Therefore,

$\begin{align} f(1)-f(0)&=\int_0^1 f^\prime(s)\,ds\\ &=\int_0^1 \left(\frac{\ln(1-s)}{s}-\frac{\ln(1+s^2)}{2s}\right)\,ds\\ -\frac{\pi^2}{16}-\frac{\pi^2}{8}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\ -\frac{3\pi^2}{16}&=J-\int_0^1 \frac{\ln(1+s^2)}{2s}\,ds\\ \end{align}$

In the latter integral perform the change of variable $y=s^2$,

$\begin{align} -\frac{3\pi^2}{16}&=J-\frac{1}{4}\int_0^1 \frac{\ln(1+y)}{y}\,dy\\ &=J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)-\ln(1-y)}{y}\,dy\\ &=J+\frac{1}{4}J-\frac{1}{4}\int_0^1 \frac{\ln(1-y^2)}{y}\,dy\\ \end{align}$

In the latter integral perform the change of variable $x=y^2$,

$\begin{align} -\frac{3\pi^2}{16}&=J+\frac{1}{4}J-\frac{1}{4}\times \frac{1}{2}J\\ &=\frac{9}{8}J\\ \end{align}$

Therefore,

$\begin{align}J&=\frac{8}{9}\times -\frac{3}{16}\pi^2\\ &=\boxed{-\frac{\pi^2}{6}}\end{align}$

I assume you're not happy with $$ \ln(1-x) = -\sum_{n=1}^\infty \frac{x^n}{n}, \qquad x\in(-1,1) $$ from which $$\begin{align} \int_0^1 \frac{\ln(1-x)}{x}dx &= -\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n} dx = -\int_0^1 \sum_{n=0}^\infty \frac{x^{n}}{n+1} dx \\&\stackrel{\rm (\ast)}{=} -\sum_{n=0}^\infty \frac{1}{n+1}\int_0^1 x^n dx = -\sum_{n=0}^\infty \frac{1}{(n+1)^2}\\ &= -\sum_{n=1}^\infty \frac{1}{n^2} = \boxed{-\frac{\pi^2}{6}} \end{align}$$ ? (It's not Feynman's trick, just a nice series representation for $\ln(1-x)$ which goes a long way.)

The only "catch" here is that swapping $\int$ and $\sum$ in $(\ast)$ actually requires a little bit of justification.