Find all functions from $\mathbb{R}-\{0,1\}$ to $\mathbb{R}$ such that

$$
f(x)+f\left(\frac{1}{1-x}\right)=x
$$
The functional equation is expressed with the values $x, \frac{1}{1-x}$ and $\frac{1}{x}-1$, which gives:

$$
\begin{aligned}
& f(x)+f\left(\frac{1}{1-x}\right)=x \\
& f\left(1-\frac{1}{x}\right)+f\left(\frac{1}{1-x}\right)=x \\
& f(x)+f\left(1-\frac{1}{x}\right)=1-\frac{1}{x}
\end{aligned}
$$

By summing the first and third lines and then subtracting the second, we find

$$
f(x)=\frac{1}{2}\left(x+1-\frac{1}{x}-\frac{1}{1-x}\right)
$$