$f(a)=f(-1)+f(-a)=2f(-1)+f(a)$, hence $f(-1)=0$. Therefore $f(-a)=f(a)$, and $f(1)=0$. $f\equiv 0$ satisfies the condition. Now suppose not: Let $p>0$ be smallest ineteger such that $f(p)\neq 0$. If $p$ is not a prime, $p=ab$, and $f(p)=f(a)+f(b)$ so at least one oth $f(a),f(b)$ is not 0, contadicts to minimality. Hence $p$ is a prime. Since $f(p)>0$, for all $k$ $f(pk)>0$ holds. Then if $0<r<p$, $f(r)=0\geq \min(f(kp+r),f(kp))$, which means $f(kp+r)=0$. Now it's easy to show $f(n)=v_{p}(n)$. Therefore $f(x)=0$ or $f(x)=v_{p}(x)$ for some prime $p$.

IsoLyS wrote: Therefore $f(x)=0$ or $f(x)=v_{p}(x)$ for some prime $p$. Precisely : $f(x)=kv_p(|x|)$ for some $k\in\mathbb N_0$ and some prime $p$