Solution:

Let $x \in \mathbb{R} \backslash\{0,1\}$ and $y=1-\frac{1}{x}$ and $z=\frac{1}{1-x}$. It is easy to see that together with $x$, $y$ and hence also $z$ belong to $\mathbb{R} \backslash\{0,1\}$. Substituting $y$ and $z$ into the original equation leads to:
$$f\left(1-\frac{1}{x}\right)+f\left(\frac{1}{1-x}\right)=2-\frac{1}{x} \text{ and } f\left(\frac{1}{1-x}\right)+f(x)=1+\frac{1}{1-x}$$
Subtracting the last two relations gives:
$$f(x)-f\left(1-\frac{1}{x}\right)=\frac{1}{1-x}+\frac{1}{x}-1$$
and adding this to the original equation finally leads to
$$f(x)=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{x}+x\right)=\frac{-x^{3}+x^{2}+1}{2 x(1-x)}$$