IMO Hk TST

The only solution is $f(x) = 3x$.
When $f(x) = 3x$, we have
$$f(3x - 1 - f(x)) = f(-1) = -3 = 3(-1) = 3(f(x) - 1 - 3x).$$
Also, $\frac{f(x)}{x} = 3$ is a constant. So $f(x) = 3x$ is a solution.

Now, we show that this is the only solution. Let $g(x) = 3x - 1 - f(x)$ for any $x \in \mathbb{R}$. If $g(x) = 0$ and $x \neq 0$, then $f(x) = 3x - 1$ and so $\frac{f(x)}{x} = 3 - \frac{1}{x}$. By condition (i), there are finitely many $x$ such that $g(x) = 0$.
For those $x \in \mathbb{R}$ with $g(x) \neq 0$, we rewrite condition (ii) as follows:
$$\frac{f(g(x))}{g(x)} = \frac{3(f(x) - 1 - 3x)}{3x - 1 - f(x)} = -3 - \frac{6}{3x - 1 - f(x)} = -3 - \frac{6}{g(x)}.$$
By condition (i), the left-hand side only takes finitely many values. This implies $g(x)$ only takes finitely many values (possibly 0).
Consider any $y \in \mathbb{R}$ such that $|f(y) - 3y| = |g(y) + 1|$ is maximized. For this $y$, we use (ii) to obtain
$$f(g(y)) - 3g(y) = 3(f(y) - 1 - 3y) - 3(3y - 1 - f(y)) = 6(f(y) - 3y).$$
This implies $|f(g(y)) - 3g(y)| = 6|f(y) - 3y|$. By the choice of $y$, we must have $f(y) - 3y = 0$. Since $y$ is chosen to maximize $|f(y) - 3y|$, we can only have $f(x) = 3x$ for any $x \in \mathbb{R}$.