Look here: https://artofproblemsolving.com/community/c6h1074466p4686261

We claim $f=0$ is the only solution. $P(y,1)$ implies $f(x)\ge 0$ and thus $f(x)y+f(y)\ge 0$, then making $y=-z$ we get that $f$ is limited, as $f(-z)\ge f(x)z$ and if $f$ was not limited we could make $f(x)$ as big as we want which is a contradiction as $z$ is fixed. Now there is $c>f(x)$ for all $x$. But then $f(xy)\le cy+c$, and making $y=-1$ we get that $0 \ge f(-x)$ thus $f(x)=0$ for all $x$ as $-x$ covers all real numbers.