Here are the properties of $f(x)$ I identified so far. I doubt it's enough to get any credit on a math olympiad, but anyway...
- The value of $f(1)$:
$$\begin{align}&x=y=1 & \Rightarrow \quad& f(1)+f(1) \ge f(1), & f(1) \ge 0 \\ &x=y=-1 & \Rightarrow \quad& -f(-1) + f(-1) \ge f(1), & f(1) \le 0 \end{align}$$
$$\therefore f(1) = 0$$
- A relationship between $f(x)$ and $f(-x)$, and some knowledge about $f(0)$
$$yf(x) + f(y) \ge f(xy) \\ -yf(x) + f(-y) \ge f(-xy) \\ \therefore f(y)+f(-y) \ge f(xy) + f(-xy)$$ $$xyf\left(\frac{1}{x}\right) + f(xy) \ge f(y) \\ -xyf\left(\frac{1}{x}\right) + f(-xy) \ge f(-y) \\ \therefore f(xy)+f(-xy) \ge f(y) + f(-y), \text{ given } x\ne 0$$
Therefore the two sides are equal. Since any two sums in the form of $f(x)+f(-x)$ ($x$ nonzero) are equal by above, we can say that it equals some constant $C$. $$\exists C \in \mathbb{R}, \; \forall x \ne 0, \; f(x)+f(-x)=C$$
Since $f(1) = 0$, $f(-1) = C$ holds, and we can simplify the above to $$ \forall x \ne 0, \; f(x)+f(-x)=f(-1)$$
The first inequality still holds for $x=0$, so $$ f(0) \le \frac{C}{2}$$ (we can't say they are equal because $f$ is not necessarily continuous.)