Iranian Mathematical Olympiad

Plugging $y = 0$ it follows that $f(f(x)) = x$, hence the function is bijective. Plugging $(x, y) = (0, 1)$ it follows that $f(f(0) + f(1)) = 1 = f(f(1))$ using injectivity to obtain $f(0) = 0$. Now, taking $f$ from both sides and using $f(f(x)) = x$ to obtain $f(x) + y f(y) = f(x + |y|^2)$. Plugging $x = 0$ yields $y f(y) = f(|y|^2)$. Hence, $f(x) + f(z) = f(x + z)$ where $z = |y|^2$ is non-negative real number. Hence, plugging $(x, y) = (0, f(y))$ gives $y f(y) = f(|f(y)|^2) = f(|y|^2)$. Again, using injectivity, we have $|f(y)| = |y|$. Let $x, z > 0$ be real numbers then
$$x + z = |f(x)| + |f(z)| \ge |f(x) + f(z)| = |f(x+z)| = x + z$$
Hence, the equality case of the triangle inequality occurs. Thus $\frac{f(x)}{f(z)} > 0$. Let $C = f(1)$ it follows that $f(x) = Cx$ for all $x > 0$. Now, by $y f(y) = f(|y|^2) = |y| f(|y|) = C|y|^2$. We find that $f(y) = C \frac{|y|^2}{y} = C \bar{y}$.

Plugging it into the original functional equation, we obtain,
$$f(f(x) + y f(y)) = \overline{C \cdot C \bar{x} + C y \bar{y}} = |C|^2(x + |y|^2) = x + |y|^2$$
Thus, $|C| = 1$.

After we obtained $|f(y)| = |y|$, we can say $x + |y|^2 \ge |f(x) + y f(y)| = x + |y|^2$. Hence, $\frac{f(x)}{y f(y)} > 0$. The rest would be analogous to the previous proof.