Taking $x=y=0\implies f(0) = f(0)+f(0)\implies f(0)=0$. Let's prove $f(kx)=kf(x), k\in\mathbb{N}, x\in\mathbb{Q}$ by induction. This is true for $k=1$ Assuming it is true for $k$, then for $k+1$: $$f((k+1)x)=f(kx+x)=f(kx)+f(x)=kf(x)+f(x)=(k+1)f(x)$$ Taking $y=-x$, we have $0=f(x)=f(x+(-x))=f(x)+f(-x)$. This implies $f(-x)=-f(x)$. Therefore $f(-kx)=-kf(x)$ holds. Therefore $f(kx)=kf(x)$ holds for all $k\in\mathbb{Z}, x\in\mathbb{Q}$. Taking $y=1/x$ then $f(1)=f(k(1/l))=kf(1/k)\implies f(1/k)=(1/k)f(1)$. For $m\in\mathbb{Z}$ and $n\in\mathbb{N}$, $f(m/n)=mf(1/n)=(m/n)f(1)$. Hence $f(x)=cx$ where $c=f(1)$.