Functional equation: $f\left(\frac{x-1}{x}\right)+ f\left(\frac{1}{1-x}\right)= 2- 2x$ There is a function given $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x
,f\colon \Bbb R\setminus\{0,1\}\to \Bbb R$ How many fuction exist? I have no idea how to start
$f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x\tag{1}$

Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(1)$

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

Similarly Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(2)$

$\displaystyle f(x)+f\left(\frac{x-1}{x}\right) = \frac{2x}{x-1}\tag{3}$

Now $(1)-(2)-(3),$ we get

$$\displaystyle -2f(x) = 2-2x-\frac{2}{x}-\frac{2x}{x-1} = -2\left(\frac{x^3+x-1}{x^2-x}\right)$$

So $\displaystyle f(x) = \frac{x^3+x-1}{x^2-x}$ for all $x\in \mathbb{R}\setminus\{0,1\}$
Hint: If you let $g(x)=\dfrac{x-1}x$, then $g^{-1}(x)=\dfrac 1{x-1}$.

So $f(g(x))+f(g^{-1}(x))=2-2x$.