Austrian Mathematical Olympiad

The function $f$ is injective using the variable $x$ (on the left $x$ only occurs as $f(x)$, on the right $x$ is free with a non-vanishing factor, so substituting $x=a$ and $x=b$ with $f(a)=f(b)$ gives the desired conclusion).

We set $x=0$ and remove the outer $f$ due to the injectivity and obtain $f(y)=y+C$.

Substituting into the original equation shows that this is equivalent to $\alpha =\beta$ (coefficient of $x$) and $(1+\alpha )C=0$ (constant coefficient).

This gives the solutions $f(x)=x$ for $\alpha =\beta$ and $f(x)=x+C$ for $\alpha =\beta =-1$.