Croatian Mathematical Competitions

Setting $x = y$ gives $0 = f(0)(f(2x) - 1)$, $\forall x \in \mathbb{R}$.

First case: $f(0) \ne 0$
We have $f(2x) - 1 = 0$, $\forall x \in \mathbb{R}$, so $f(x) = 1$, $\forall x \in \mathbb{R}$. We check that this function is a solution.

Second case: $f(0) = 0$
Setting $y = 0$ gives $f(f(x)) - f(x) = f(x)(f(x) - 1)$, i.e.
$$f(f(x)) = (f(x))^2, \quad \forall x \in \mathbb{R}. \qquad (\star)$$
Interchanging $x$ and $y$ gives $f(x+f(y)) - f(y+f(x)) = f(y-x)(f(x+y)-1)$, which added to the starting equation gives
$$0 = (f(x+y) - 1)(f(x-y) + f(y-x)), \quad \forall x, y \in \mathbb{R}.$$
Setting $y = -x$ in thus obtained equation gives $0 = (f(0) - 1)(f(2x) + f(-2x))$, so $f(0) = 0$ implies $f(2x) = -f(-2x)$, $\forall x \in \mathbb{R}$ and hence the function $f$ is odd.

Now we conclude
$$f(f(x)) \stackrel{*}{=} (f(x))^2 = (-f(-x))^2 = (f(-x))^2 \stackrel{*}{=} f(f(-x)) = -f(-f(-x)) = -f(f(x)), \qquad (\star\star)$$
for every $x \in \mathbb{R}$

Finally, $(\star\star)$ and $(\star)$ give $2f(f(x)) = 0$, so $2(f(x))^2 = 0$. We check directly that $f(x) = 0$, $\forall x \in \mathbb{R}$ is also a solution.