(APMO 2011)

Find all bounded real functions $f$ such that for all real numbers $x, y$, we have

$$
f(x f(y)) + y f(x) = x f(y) + f(x y)
$$
By setting $x=0$, we find $f(0)=0$. Then by setting $y=1$, we have $f(x f(1))=x f(1)$, and since $f$ is bounded, $f(1)=0$.

Now, with $x=1$ we get $f(f(x))=2 f(x)$, so $f^{n}(x)=2^{n-1} f(x)$. Again, since $f$ is bounded, we have $f(x) \leqslant 0$ for all $x$. Suppose there exists $r$ such that $f(r) < 0$. Since $f(x) / x \leqslant 0$, we have $f(x)=0$ (since $f$ cannot be zero on $\mathbb{R}^{-*}$). By testing the pair $(-1, x)$, we see that $f(-x)=x f(-1)$, so with $-x=f(r)$, $f(f(r))=-f(r) f(-1)$, but $f(f(r))=2 f(r)$. Thus, $f(-1)=-2$. Therefore:

$f(x)=-2 x$ if $x<0$, $f(x)=0$ otherwise. It is easy to verify that this function and the zero function are solutions.