Mongolian National Mathematical Olympiad

Answer: $f(x) = Ax + B$ for real numbers $A$ and $B$.
For $x \in \mathbb{R}$, let $a = \frac{x-1}{2}$, $b = \frac{x+1}{2}$ and $c = \frac{1-x}{2}$. Then we have
$$\begin{cases} a+b=x \\ b+c=1 \\ c+a=0 \end{cases}$$
and
$$\begin{cases} a-b=-1 \\ b-c=x \\ c-a=1-x \end{cases}$$
, therefore we have $-f(x) + x f(1) + (1-x) f(0) = 0$. Denoting $A = f(1) - f(0)$ and $B = f(0)$, we get $f(x) = Ax + B$.

Conversely, for any $A$ and $B$, the function $f(x) = Ax + B$ satisfies
$$(a-b)f(a+b) + (b-c)f(b+c) + (c-a)f(c+a) = \sum_{\text{cyc}} A(a^2 - b^2) + B(a-b) = 0.$$