IMO Team Selection Contest

By choosing $(x,y)=(z,1)$ in the original equation, where $z$ is any real number, we get $f(z+f(z))=2zf(1)$. However, by choosing $(x,y)=(1,z)$ in the original equation, we get $f(z+f(z))=2f(z)$. From these equations together we obtain $f(z)=f(1)z$.

By choosing $z=1$ in equation $f(z+f(z))=2zf(1)$, we get $f(1+f(1))=2f(1)$, however by choosing $z=1+f(1)$ in equation $f(z)=f(1)z$, we get $f(1+f(1))=f(1)(1+f(1))$. Altogether we get $(1+f(1))f(1)=2f(1)$ or, equivalently, $f(1)(f(1)-1)=0$, from which $f(1)=0$ or $f(1)=1$.

Therefore by $f(z)=f(1)z$ either $f(z)=0$ or $f(z)=z$ for every $z$. We can check that both functions indeed satisfy the original equation.

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Alternative solution.

By choosing $y=0$ we get $f(f(0))=2xf(0)$. As $x$ can be any real number, this can hold only if $f(0)=0$.

By choosing $x=\frac{1}{y}$, we end up with $f(1+f(1))=\frac{2}{y}f(y)$. Hence for each $y\ne 0$ the equation $f(y)=\frac{f(1+f(1))}{2}y$ holds. In the case of $f(1+f(1))=0$ this equation gives $f(y)=0$ for every $y$. We get the same result if $1+f(1)=0$, because $f(0)=0$.

However, if $1+f(1)\ne 0$ and $f(1+f(1))\ne 0$, then we can take $y=1+f(1)$ in the equation above and divide both sides by $f(1+f(1))$; we obtain $\frac{1+f(1)}{2}=1$, which implies $f(1)=1$. Now by setting $y=1$ in the equation above we get $\frac{f(1+f(1))}{2}=1$, therefore $f(y)=y$ for every $y$.