1. Plugging $y = x + f(x)$ to obtain $f(f(x + f(x))) = 2x$. Hence, the function is surjective and there is a real number $r$ such that $f(r) = 0$. Plug $x = r$ to obtain
$$f(y) = -2r + f(f(y)).$$
For each real number $z$ there is a real number $y$ such that $f(y) = z$, therefore
$$z = -2r + f(z). \text{ It follows that}$$
$$f(z) = z + 2r, \quad z \in \mathbb{R}.$$
By plugging this into the original equation one can conclude that $r = 0$ and $f(z) = z$, for all real numbers $z$.