Let's determine all functions $f: \mathbb{R} \backslash\{0,1\} \rightarrow \mathbb{R}$ that satisfy the equation

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

relation.
Solution. For any $x_{0} \in \mathbb{R} \backslash\{0,1\}$,

$$
f\left(\frac{x_{0}-1}{x_{0}}\right)+f\left(\frac{1}{1-x_{0}}\right)=2-2 x_{0}
$$

Now, substitute $\frac{1}{1-x_{0}} \in \mathbb{R} \backslash\{0,1\}$ for $x$ in the original equation:

$$
\begin{aligned}
f\left(\frac{\frac{1}{1-x_{0}}-1}{\frac{1}{1-x_{0}}}\right) & +f\left(\frac{1}{1-\frac{1}{1-x_{0}}}\right)=f\left(1-1+x_{0}\right)+f\left(\frac{1-x_{0}}{1-x_{0}-1}\right)= \\
& =f\left(x_{0}\right)+f\left(\frac{x_{0}-1}{x_{0}}\right)=2-2 \cdot \frac{1}{1-x_{0}}
\end{aligned}
$$

Now apply the substitution $x=\frac{x_{0}-1}{x_{0}}$.

$$
\begin{gathered}
f\left(\frac{\frac{x_{0}-1}{x_{0}}-1}{\frac{x_{0}-1}{x_{0}}}\right)+f\left(\frac{1}{1-\frac{x_{0}-1}{x_{0}}}\right)=f\left(\frac{x_{0}-1-x_{0}}{x_{0}-1}\right)+f\left(\frac{x_{0}}{x_{0}-x_{0}+1}\right)= \\
=f\left(\frac{1}{1-x_{0}}\right)+f\left(x_{0}\right)=2-2 \cdot \frac{x_{0}-1}{x_{0}}
\end{gathered}
$$

Subtract the original equation (1) from the sum of equations (2) and (3), then divide by 2 and replace $x_{0}$ with $x$ in the equation.

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

Substituting this function into the original functional equation, it can be verified that this function is indeed a solution:

$$
\begin{aligned}
& f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-x}\right)= \\
= & \frac{x-1}{x}+\frac{x}{x-1}+\frac{1}{\frac{x-1}{x}-1}+\frac{1}{1-x}+(1-x)+\frac{1}{\frac{1}{1-x}-1}= \\
= & \frac{x-1}{x}+1-x+(1-x)+\frac{1-x}{x}=2-2 x .
\end{aligned}
$$