Suppose that $f$ is a solution. Let
$$a = \frac{1}{2}(f(1) - f(-1)), \quad b = \frac{1}{2}(f(1) + f(-1))$$
and $g(x) = f(x) - a x - b x^2$. 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
$$\begin{aligned} (x+1)g(x) &= (x-1)g(x+1) \\ xg(x+1) &= (x+2)g(x). \end{aligned}$$
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 $f(x) = a x + b x^2$. We can check directly that any function of this form (for some $a, b \in \mathbb{R}$) satisfies the given equation.