By choosing $(x, y) = (z, 1)$ in the original equation, where $z$ is any real number, we get $f(z + f(z)) = 2z f(1)$. However, by choosing $(x, y) = (1, z)$ in the original equation, we get $f(z + f(z)) = 2 f(z)$. From these equations together we obtain $f(z) = f(1) z$.

By choosing $z = 1$ in equation $f(z + f(z)) = 2z f(1)$, we get $f(1 + f(1)) = 2 f(1)$, however by choosing $z = 1 + f(1)$ in equation $f(z) = f(1) z$, we get $f(1 + f(1)) = f(1)(1 + f(1))$. Altogether we get $(1 + f(1)) f(1) = 2 f(1)$ or, equivalently, $f(1)(f(1) - 1) = 0$, from which $f(1) = 0$ or $f(1) = 1$.

Therefore by $f(z) = f(1) z$ either $f(z) = 0$ or $f(z) = z$ for every $z$. We can check that both functions indeed satisfy the original equation.

By choosing $y = 0$ we get $f(f(0)) = 2x f(0)$. As $x$ can be any real number, this can hold only if $f(0) = 0$.

By choosing $x = \frac{1}{y}$, we end up with $f(1 + f(1)) = \frac{2}{y} f(y)$. Hence for each $y \neq 0$ the equation $f(y) = \frac{f(1 + f(1))}{2} y$ holds. In the case of $f(1 + f(1)) = 0$ this equation gives $f(y) = 0$ for every $y$. We get the same result if $1 + f(1) = 0$, because $f(0) = 0$.

However, if $1 + f(1) \neq 0$ and $f(1 + f(1)) \neq 0$, then we can take $y = 1 + f(1)$ in the equation above and divide both sides by $f(1 + f(1))$; we obtain $\frac{1 + f(1)}{2} = 1$, which implies $f(1) = 1$. Now by setting $y = 1$ in the equation above we get $\frac{f(1 + f(1))}{2} = 1$, therefore $f(y) = y$ for every $y$.