Singapore Mathematical Olympiad (SMO)

Suppose that $f$ is a solution. Let $$a=\frac{1}{2}(f(1)-f(-1)),b=\frac{1}{2}(f(1)+f(-1))$$ and $g(x)=f(x)-ax-b{x}^2$. Then $$(x+y)(g(x)-g(y))=(x-y)g(x+y)$$ and $g(1)=g(-1)=0$. Letting $y=1$ and $y=-1$ above give $$\begin{array}{rl}(x+1)g(x)& =(x-1)g(x+1)\\ xg(x+1)& =(x+2)g(x).\end{array}$$ Thus $$x(x+1)g(x)=x(x-1)g(x+1)=(x-1)(x+2)g(x)$$ for all $x$. So $g(x)=0$ for all $x$. Hence $f(x)=ax+b{x}^2$. We can check directly that any function of this form (for some $a,b\in \mathbb{R}$) satisfies the given equation.