Selection Examinations for the IMO, 2012

If we substitute $x = 1$ into the functional equation, we get
$$f(y) + f(1)^2 f(y) = f(yf(1)) + yf(1). \quad (1)$$
We successively substitute $1$, $f(1)$ and $f(1)^2$ instead of $y$ into this equation to get
$$f(f(1)) = f(1)^3, \tag{2}$$
$$f(1)^3 + f(1)^5 = f(f(1)^2) + f(1)^2, \quad (3)$$
$$f(1)^7 + 2f(1)^5 - f(1)^4 - f(1)^2 = f(f(1)^3). \quad (4)$$
In the last two equations, we have already considered (2), and in the last equation, we have also considered (3). If we now substitute $y = 1$ and $x = f(1)$ into the functional equation and use (2), we get
$$f(1)^2 + f(1)^7 = f(f(1)^3) + f(1)^4. \quad (5)$$
From (4) and (5) we derive $f(1)^5 = f(1)^2$. Hence $f(1) = 0$ or $f(1) = 1$. If $f(1) = 0$, then from (1) we get $f(y) = f(0)$ for any $y$, and we conclude that $f$ is a constant function. The only constant function that satisfies the equation is the zero function.
If, on the other hand, $f(1) = 1$, then from (1) we get $f(y) = y$ for any $y$, and we verify that this function satisfies the functional equation. The solutions are $f(x) = 0$ and $f(x) = x$.