A function $f$ is given by $f(x)+f\left(1-\frac{1}{x}\right)=1+x$ for $x \in \mathbb{R} \backslash\{0,1\}$.
Determine a formula for $f$.
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 thus 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 results in

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