Let $P(x,y)$ denotes the given assertion. Note that, $P(x+1,x-1)$ gives $x (f(x+1)-f(x-1))=af(2x)$; $P(x+1,-x)$ gives $f(x+1)-f(-x)=a(2x+1)f(1)$, and finally $P(-x,x-1)$ gives $f(-x)-f(x-1)=a(2x-1)f(-1)$. Summing the last two findings, we obtain $f(x+1)-f(x-1)=a(2x+1)f(1)+a(2x-1)f(-1)$. Inserting this into the first equation, we conclude \[ af(2x)=ax\left((2x+1)f(1)+(2x-1)f(-1)\right). \]Now, suppose $a\ne 0$. In this case, we conclude $f(2x)=x(2x+1)f(1)+x(2x-1)f(-1)$. Setting $x=r/2$, we find that there exists $\alpha,\beta$ such that $f(x)=\alpha x^2+\beta x$ for every $x\in\mathbb{R}$. Inserting this, we obtain \[ (x-y)(x+y)(\alpha(x+y)+\beta)=a(x+y)(x-y)(\alpha(x+y)+\beta). \]If $a=1$, we conclude that any such quadratic $f$ works. If $a\ne 0,1$, then taking $x\ne \pm y$, we obtain $(\alpha(x+y)+\beta)(1-a)=0$. Since $a\ne 1$, it must hold $\alpha(x+y)+\beta=0$. This is impossible if $\alpha,\beta\ne 0$. In this case, $f=0$ is the only possibility. Finally, suppose $a=0$. In this case I claim $f$ is constant. To show this, suppose first that $f$ takes three distinct values, say at points $c_1,c_2,c_3$. We then obtain, by inspecting $P(c_1,c_2)$, that $c_1+c_2=0$. Similarly, $P(c_2,c_3)$ yields $c_2+c_3=0$. Consequently, $c_1=c_3$, a contradiction. Thus, $f$ can take at most two distinct values. Now, suppose that $f(x)\in\{u,v\}$ for every $x\in\mathbb{R}$. Set $S_1=\{x:f(x)=u\}$ and $S_2=\{x:f(x)=v\}$. Now, one of these sets, say $S_1$, has cardinality at least two. We show $S_2=\varnothing$ in this case. Assume otherwise, and let $m,n\in S_1$ distinct; and $k\in S_2$. Then, $m+k=0$ and $n+k=0$, yielding $m=n$, not possible. Hence, $S_2=\varnothing$; and thus $f$ is constant in this case.