Belarusian Mathematical Olympiad

Let the function $f: \mathbb{R} \to \mathbb{R}$ satisfy the equality
$$f(f(x)) = x f(x) + x - 1 \quad (*)$$
for all $x \in \mathbb{R}$.

a) Set $c = f(0)$. We have $f(c) = f(f(0)) = 0 \cdot f(0) + 0 - 1 = -1$. So
$$-1 = f(c). \quad (1)$$
Therefore, taking into account $(*)$ for $x = c$ and (1), we obtain the required value
$$f(-1) = f(f(c)) = c \cdot f(c) + c - 1 = -c + c - 1 = -1.$$

b) Let $a = f(1)$. Taking into account $(*)$ for $x = 1$, we obtain
$$f(a) = f(f(1)) = 1 \cdot f(1) + 1 - 1 = a.$$
Since
$$a = f(a), \quad (2)$$
it follows that
$$f(a) = f(f(a)) = a \cdot f(a) + a - 1 = a^2 + a - 1. \quad (3)$$
From (2) and (3) it follows that $a^2 + a - 1 = a$, i.e. either $a = -1$ or $a = 1$.

Both the values are achieved. For example,
$$f(x) = -1 \quad \text{and} \quad f(x) = \begin{cases} -1, & \text{if } x \neq 1, \\ 1, & \text{if } x = 1. \end{cases} \quad (3)$$
It is easy to verify that both the functions satisfy the initial equation $(*)$ for all real $x$.

Some more examples:
$$f(x) = \begin{cases} \frac{1}{x}, & \text{if } x \neq 0, \\ -1, & \text{if } x = 0 \end{cases} \quad \text{or} \quad f(x) = \begin{cases} -1, & \text{if } x \neq 0, \\ \alpha, & \text{if } x = 0, \end{cases}$$
where $\alpha$ is any real number different from $0$.
Another function: $f(x) = \frac{1}{x}$, if $x \in A$, and $f(x) = -1$ for all other $x$, where $A$ is an arbitrary subset of $\mathbb{R} \setminus \{0\}$, such that $A^{-1} \subset A$.