Answer: $f(x) = a$, $g(x) = a - a x$ for all $a \in \mathbb{R}$.

Let $a = f(0)$, $b = g(0)$. Note that due to the symmetry the given equation implies
$$x f(y) + g(x) = y f(x) + g(y). \tag{1}$$
Set $y = 0$ in (1), then we have $g(x) = b - a x$. Now set $y = 1$ in (1), then $f(x) = (f(1) - a)x + a = a + c x$. Now substituting $g(x)$ and $f(x)$ in the given equation we easily obtain that $c = 0$, $b = a$.