, 2012

Put $x = y = 0$; we get $f(0) = 3f(0)$ so that $f(0) = 0$. Taking $y = -1$, we get $f(-1) = f(x) + f(-1) + f(-x)$ so that $f(-x) = -f(x)$ for all $x \in \mathbb{R}$. Taking $y = 1$, we get $f(2x + 1) = 2f(x) + 1$, for all $x \in \mathbb{R}$. Replace $x$ by $u + v + uv$ in this, we get
$$f(2(u + v + uv) + 1) = 2f(u + v + uv) + f(1) = 2f(u) + 2f(v) + 2f(uv) + f(1), \quad (1)$$

for all $u, v \in \mathbb{R}$. Taking $x = u$, $y = 2v + 1$ in the original equation, we get
$$f(u + 2v + 1 + 2uv + u) = f(u) + f(2v + 1) + f(2uv + u),$$
and this reduces to
$$f(2(u + v + uv) + 1) = f(u) + f(2v + 1) + f(2uv + u),$$
(2)
for all $u, v \in \mathbb{R}$. Comparing (1) and (2) and using $f(2x + 1) = 2f(x) + 1$, we get
$$f(2uv + u) = f(u) + 2f(uv),$$
(3)
for all $u, v \in \mathbb{R}$. Put $v = -1/2$ in (3) to get
$$0 = f(0) = f(u) + 2f(-u/2) = f(u) - 2f(u/2).$$
This shows that $f(u/2) = f(u)/2$ or $f(2u) = 2f(u)$ for all $u \in \mathbb{R}$. Thus
$$f(2uv + u) = f(2uv) + f(u),$$
for all $u, v \in \mathbb{R}$. Given any $x \neq 0$ and $y$, we can find $u, v$ such that $2uv = y$ and $u = x$. Thus $f(x+y) = f(x) + f(y)$ for all $x \neq 0$ and $y$. However this also valid for $x = 0$. Thus $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$.