How to find the $n$-th derivative of the function $\log(1-x(1-x))$ I know how to find the $n$ derivative of $\log (1-x)$ which is $-\frac{(n-1)!}{(1-x)^n}$, but how to find the $n$-th derivative of the function $\log(1-x(1-x))$?

By purely symmetry conditions we get that $x(1-x)$ is a parabola with a vertex at $x=\frac{1}{2}$ which means we can rewrite it as
$$x(1-x) = \frac{1}{4}-\left(x-\frac{1}{2}\right)^2$$
Then the log becomes
$$\log\left(\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\right)$$
Finding the $n$th derivative of this function at $\frac{1}{2}$ is equivalent to finding the $n$th derivative of
$$\log\left(\frac{3}{4}+x^2\right) = \log\frac{3}{4} + \log\left(1+\frac{4}{3}x^2\right)$$
at $0$. The function on the right has a known Taylor series
$$\log(1+x)  = \sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$$
Can you take it from here?
The answer when you want to check is

 $$f^{(n)}\left(\frac{1}{2}\right) = \begin{cases} \log\frac{3}{4} & n=0 \\ 0 & n\text{ odd}\\ \frac{(-1)^{\frac{n}{2}+1}2^{n+1}(n-1)!}{\sqrt{3^n}} & n\text{ even}\\ \end{cases}$$

