Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real $x$ different from 1:

$$
f(x)+f\left(\frac{1}{1-x}\right)=x
$$
Let $x$ be different from 0 and 1.

We apply the equation to $\frac{1}{1-x}$ instead of $x$, we then have:

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

We reapply $\frac{1}{1-x}$ instead of $x$ in this last equation:

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

We have three equations with three unknowns, so we can solve the system. We take the initial equation plus the last one from which we subtract the second to obtain:

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

Hence: $f(x)=\frac{x^{3}-2 x}{2 x(x-1)}$ for $x \in \mathbb{R}-\{0,1\}$.

For $x=0$, the initial equation implies $f(0)+f(1)=0$, which is the only constraint on $f(0)$ and $f(1)$.

Thus, the functions that are solutions (because they fit) are the $f: \mathbb{R} \longmapsto \mathbb{R}$ such that there exists $c \in \mathbb{R}$ such that $f(x)=\frac{x^{3}-2 x}{2 x(x-1)}$ for $x$ different from 0 and 1, $f(0)=c$ and $f(1)=-c$.