<s>[INST] Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x)$ has a zero at $x=\frac{1}{2}$. But can we conclude that this is unique？
 [/INST] My second solution:
It suffices to prove that, for all $x \in (0, 1/2]$,
$$\ln x \ln (1 - x) \le \ln^2 2.$$
Let $f(x) = \ln x \ln (1 - x) - \ln^2 2$. We have
$$f'(x) = \frac{(1 - x)\ln (1 - x) - x\ln x}{x(1 - x)}.$$
Since $u\mapsto \ln u$ is concave on $(0, \infty)$, we have  (see the remarks)
$$\frac{1 - 2x}{1 - x}\ln \frac{1}{x} + \frac{x}{1 - x}\ln 1 \le \ln\left(\frac{1 - 2x}{1 - x} \cdot \frac{1}{x} + \frac{x}{1 - x} \cdot 1\right) = \ln\frac{1 - x}{x}$$
which results in $(1 - x)\ln(1 - x) - x\ln x \ge 0$.
Thus, $f'(x) \ge 0$. Also, $f(1/2) = 0$.
Thus, $f(x) \le 0$ for all $x\in (0, 1/2]$.
We are done.

Remarks: i) Here we used @trancelocation's nice idea in How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$?.
ii) Alternatively, using @Quanto's nice idea in the link above, we have
\begin{align*}
 (1 - x)\ln(1 - x) - x\ln x
 = x\int_x^{1/2} \frac{1}{u^2}\int_0^u \frac{1 - 2v}{1 - v} \mathrm{d} v \mathrm{d}u \ge 0.
\end{align*}
</s>