Mr. SBM?

You better call for usjl. The only problem here is limiting right side for negative arguments of the function. BTW. I believe arqady solved $\definecolor{A}{RGB}{250,120,0}\color{A}\fbox{A3}$ (or generally knows how the proof will go) while we're still seating on $\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{A1.}$

Put $x=y=z$ to see that $f(x)=x^2$ for $x \ge 0$. Moreover, putting $y=z=-x$ we see that $f(x) \le x^2$ for all $x$. From now on, we can ignore the left inequality and only work with the right one. Substituting $xy=a, yz=b, zx=c$ we get that $f(a)+f(b)+f(c) \ge ab+ac+bc$ whenever $abc>0$. So choosing $a,b<0, c>0$ we see that $f(a)+f(b) \ge ab+(a+b)c-c^2$. Taking the limit $c \to 0$ we see that $f(a)+f(b) \ge ab$ whenever $a,b<0$. In particular, on $a=b$ we get that $f(a) \ge \frac{a^2}{2}$. But it's easy to check that any function $f$ with $f(x)=x^2$ for $x \ge 0$ and $f(x) \in \left[\frac{x^2}{2}, x^2\right]$ for $x<0$ satisfies the condition.