Romanian Mathematical Olympiad

We claim that the solutions are $f_1(x) = x$ and $f_2(x) = -2x/3$, for all real $x$.

Set $x = y - 3f(y)$ to obtain $f(y - 3f(y)) = -2y$, $y \in \mathbb{Q}$. Replacing $y$ with $y - 3f(y)$ in the initial equation gives $f(x - 6y) = f(x) - 2y + 2(y - 3f(y))$, hence $f(x - 6y) = f(x) - 6f(y)$, for all $x, y \in \mathbb{Q}$.

Set now $x = y = 0$ to get $f(0) = 0$ and replace $x = 6y$ to obtain $f(6y) = 6f(y)$, for all $y \in \mathbb{Q}$.

We derived that $f(x - 6y) = f(x) - f(6y)$, which, for $u = 6y$ and $v = x - 6y$, yields $f(u + v) = f(u) + f(v)$, for all $u, v \in \mathbb{Q}$. The solution of this classical functional equation is $f(x) = x f(1)$, $x \in \mathbb{Q}$.

On the other hand, by setting $x = y = 1$ in the initial equation we obtain $3f^2(1) - f(1) - 2 = 0$, hence $f(1) = 1$ or $f(1) = -2/3$, thus proving our claim.