Solution:
Let $P(x, y)$ be the problem statement. Note that if $f(a)=f(b)$ where $a, b \in \mathbb{R}$, then $P(x, a)$ and $P(x, b)$ imply $a=b$. Thus $f$ is injective.

Then $P(x,-f(x))$ implies $f(2 f(x))=f(x-f(-f(x)))$, so by injectivity we obtain $f(-f(x))=x-2 f(x)$.

Now $P(x, 0)$ gives $f(2 f(x))=f(x-f(0))+f(x)$. Combining this with $P(x, y)$, we obtain $f(x-f(0))=f(x-f(y))+y$.

Setting $x=0$ here yields $f(-f(0))=f(-f(y))+y$, and so $f(-f(0))=2 y-f(y)$. In particular, this implies $f(y)=y+c$ for some constant $c$.

Substituting this into $P(x, y)$, we obtain
$$2 x+3 c=x-y+x+y+c$$
which implies that $c=0$. It is easy to see that $f(x)=x$ works and is therefore the only solution.