XVIII-th Macedonian mathematical Olympiad

If we choose $x = 0$, then
$$f(y f(0)) = f(f(0)).$$
If $f(0) \ne 0$, then for $y = \frac{1}{f(0)}$, we get
$$f(t) = f(f(0)) = c,$$
i.e. $f$ is a constant function. But then, substituting in (1), the equation gets the form
$$c = c + c x, \quad x \in \mathbb{R}.$$
As $x$ is arbitrary, we get that $c = 0$. Due to the resulting contradiction, $f(0) = 0$.

II. Suppose that $f(x) = 0$ for some $x \ne 0$. Then, substituting in the initial equation
$$x f(y) = 0,$$
for each $y \in \mathbb{R}$. Therefore, $f(y) = 0$ for each $y \in \mathbb{R}$. Due to the resulting contradiction, $f(x) = 0$, $x \in \mathbb{R}$.

It is not hard to check that $f(x) = 0$, $x \in \mathbb{R}$ is a solution of the initial equation.

III. If $y = 0$, then $f(x) = f(f(x))$. Now, substituting in the first equation,
$$f(x + y f(x)) = f(x) + x f(y).$$

IV. Let $f(1) = a \neq 0$. Then
$$\begin{align*} x = 1, y = 1 &\Rightarrow f(1 + a) = 2a \\ x = 1, y = -1 &\Rightarrow f(1 - a) = a + f(-1) \\ x = 1 + a, y = -1 &\Rightarrow f(1 - a) = f(1 + a - 2a) = 2a + (1 + a) f(-1) \end{align*}$$
Now from (1) and (2) we get
$$\begin{align*} a + f(-1) &= 2a + (1 + a) f(-1) \\ -a &= a f(-1) \\ f(-1) &= -1 \\ f(1 - a) &= a - 1 \end{align*}$$
On the other hand
$$\begin{align*} x &= 1 - a, y = 1 \\ x &= 1 \\ x &= -1 \end{align*} \begin{align*} 0 &= f(1 - a + a - 1) = a - 1 + (1 - a) a = - (a - 1)^2 \\ &= a - 1 \\ &= f(1 + y) = 1 + f(y) \\ &= f(-1 - y) = -1 - f(y) \end{align*}$$
Now from (3) and (4) $-f(y) = 1 + f(-1 - y) = f(-y)$. Therefore, for $y = -1$ we have
$$f(x - f(x)) = f(x) - x.$$
Finally,
$$\begin{align*} f(x - f(x)) &= f(f(x - f(x))) = f(f(x) - x) = -f(x - f(x)) \\ f(x - f(x)) &= 0 \\ x - f(x) &= 0 \\ f(x) &= x \end{align*}$$
This function satisfies the equation, so the required solutions are: $f(x) = 0$ and $f(x) = x$.