Belarusian Mathematical Olympiad

Suppose that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$f(f(x)) = 1 - x f(x) \quad (*)$$
for any $x \in \mathbb{R}$.

Set $c = f(0)$. Using $(*)$ for $x = 0$, we obtain $f(c) = f(f(0)) = 1 - 0 \cdot f(0) = 1 - 0 \cdot c = 1$.
Since
$$1 = f(c), \quad (1)$$
we have $f(1) = f(f(c)) = 1 - c \cdot f(c) = 1 - c \cdot 1 = 1 - c$. Similarly, since
$$1 - c = f(1), \quad (2)$$
we have
$$f(1-c) = f(f(1)) = 1 - f(1) = 1 - (1-c) = c. \quad (3)$$
It follows that $f(c) = f(f(1-c))_{(*)} = 1 - (1-c) \cdot f(1-c)_{(3)} = 1 - (1-c)c = c^2 - c + 1$, i.e.
$$c^2 - c + 1 = f(c). \quad (4)$$
From (1) and (4) it follows that $c^2 - c + 1 = 1$, so either $c = 0$ or $c = 1$.
We show that both the equalities $c = 0$ and $c = 1$ give a contradiction. Indeed, if $c = 0$, then from definition of $c$ it follows that $0 = f(0)$, but (1) gives $1 = f(0)$, a contradiction. If $c = 1$, then (1) and (2) give $1 = f(1)$ and $0 = f(1)$ respectively, a contradiction. Therefore there is no function satisfying $(*)$.