Selection Examinations for the IMO

First, assume that $f(0)=0$. Inserting $x=0$ into the functional equation we get $f(y)=0$ for all $y\in \mathbb{R}$. This function satisfies the equation.

Now, let $f(0)\ne 0$. Inserting $y=0$ into the equation we get $$(x-2)f(0)+f(2f(x))=f(x)$$ for all $x\in \mathbb{R}$. Obviously, this means that $f$ is injective.

Now, put $x=2$ into the equation to get $$f(y+2f(2))=f(2+yf(2))\text{for all }y\in \mathbb{R}.$$ Since $f$ is injective, we have $y+2f(2)=2+yf(2)$ for all $y\in \mathbb{R}$. If we insert $y=0$, then we get $f(2)=1$.

Since $f(2)=1$ and $f$ is injective, we have $f(3)\ne 1$. Insert $x=3$ and $y=\frac{3}{1-f(3)}$ into the equation to get $$f\left(\frac{3}{1-f(3)}+2f(3)\right)=0.$$ We have shown that $f$ has a zero. Let $a$ be a zero of $f$, $f(a)=0$. Inserting $y=a$ into the initial equation yields $$f(a+2f(x))=f(x+af(x))\text{for all }x\in \mathbb{R}.$$ Since $f$ is injective, we have $a+2f(x)=x+af(x)$ for all $x\in \mathbb{R}$. Since $a\ne 2$ (remember that $f(2)=1\ne 0$), we get $f(x)=\frac{x-a}{2-a}$.

Finally, inserting $f(x)=\frac{x-a}{2-a}$ into the equation we get $a=1$. The only two solutions to this functional equation are the functions $f(x)=0$ and $f(x)=x-1$.