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？
Note that the derivative is
$$f’(x) = \frac{\ln(1 - x)}{x} - \frac{\ln x}{1 - x}$$
for $0 < x < 1.$ We want to show that $f(x)$ is maximized when $x = 1/2.$
Since we already know that $f’(1/2) = 0,$ it remains to show that $f’’(1/2) < 0$ and that $f’(x) \neq 0$ for all other $x$ in the interval.
We have
$$\begin{align} f’’(x) &= \frac{1}{x}\cdot \frac{-1}{1 - x} - \frac{\ln(1 - x)}{x^2} - \frac{1}{x}\cdot\frac{1}{1 - x} - \frac{\ln x}{(1 - x)^2} \\ &= -\frac{2}{x (1 - x)} - \frac{\ln(1 - x)}{x^2} - \frac{\ln x}{(1 - x)^2}, \end{align} $$
from which we can compute that $f’’(1/2) = 8 \ln 2 - 8.$ Since $\ln 2 < 1,$ it follows that $f’’(1/2) < 0.$
If we can show that the function $g$ given by $$g(x) = \frac{\ln(1 - x)}{x}$$ for $0 < x < 1$ is injective, then that will be sufficient to prove that $f’(x)$ is only $0$ for $x = 1/2.$ To do that, we will apply the inequality $$\ln x > 1 - \frac{1}{x}$$ for $x$ in this interval. If you are unfamiliar with this inequality, you can see this question.
From
$$\begin{align} g’(x) &= -\frac{1}{x(1 - x)} - \frac{\ln(1 - x)}{x^2} \\ &< -\frac{1}{x(1 - x)} - \frac{1}{x^2}\left(1 - \frac{1}{1 - x}\right) \\ &= -\frac{1}{x(1 - x)} + \frac{1}{x(1 - x)} \\ &= 0, \end{align} $$
it follows that $g$ is strictly decreasing. Thus, the only zero of $f’(x)$ occurs when $x = 1/2$.
By the second derivative test, we have $$\ln x \ln(1 - x) \leq \ln^2(1/2) = \ln^2 2.$$
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 $$f'(x) = \frac{(1 - x)\ln (1 - x) - x\ln x}{x(1 - x)}.$$ Let $g(x) = (1 - x)\ln (1 - x) - x \ln x$. We have $$g'(x) = -\ln [x(1 - x)] - 2.$$ Let $x_0 = \frac{1}{2} - \frac{1}{2}\sqrt{1 - 4\mathrm{e}^{-2}} \in (0, 1/2)$. Then, $g'(x_0) = 0$, and $g'(x) > 0$ on $(0, x_0)$, and $g'(x) < 0$ on $(x_0, 1/2]$. Also, $\lim_{x\to 0^{+}} g(x) = 0$ and $g(1/2) = 0$. Thus, $g(x) \ge 0$ for all $x \in (0, 1/2]$.
Thus, $f'(x) \ge 0$ for all $x\in (0, 1/2]$. Also, $f(1/2) = 0$. Thus, $f(x) \le 0$ for all $x\in (0, 1/2]$.
We are done.
Put $x=e^{-y}$ the define :
$$f(y)=-y\ln(1-e^{-y})$$
We have :
$$f''(y)=\frac{e^y(y-2)+2}{(e^y-1)^2}$$
We can find the zero of the second derivative using the Lambert's function and the derivative have a zero at $y=\ln(2)$ .
The conclusion follow .
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*}
A proof by someone posted here
\begin{align*} \ln x\ln(1-x)&=\sum_{n=1}^\infty\frac{1-x^n-(1-x)^n}{n^2}\\ &\le \sum_{n=1}^\infty\frac{1-2\left[\frac{x+(1-x)}{2}\right]^n}{n^2}\color{blue}{~~(\text{Jensen's inequality)}}\\ &=\sum_{n=1}^{\infty}\frac{1-\frac{1}{2^{n-1}}}{n^2}\\ &=\zeta(2)-2\rm{Li_2}\left(\frac{1}{2}\right)\\ &=\frac{\pi^2}{6}-2\left(\frac{\pi^2}{12}-\frac{\ln^2 2}{2}\right)\\ &=\ln^2 2. \end{align*}
