$(f \circ f \circ f)(x) = (f \circ f)(x) + x$

I have a very strange question in my homework, I'm not sure how I can do it:
$(f \circ f \circ f)(x) = (f \circ f)(x) + x$
Show $f$ is injective
Compute $f(0)$
I can't see how I could possible isolate f(x) here, I thought maybe I could get $(f \circ f)(x)$ with $x$, and replace the x with an unknown function $g(x)$, giving me:
$f(x) = x + g(x)$
However, even if this is right, I'm still lost on what to do here.
Any help would be appreciated, thanks.

To prove that $f$ is injective, you can proceed as usual: fix $x$ and $y$ in the domain of $f$, assume that $f(x)=f(y)$, and prove that $x=y$.
In this case, note that if $f(x)=f(y)$, then by applying $f$ we have that $(f \circ f)(x) = (f \circ f)(y)$, and by applying $f$ again, we have that $(f \circ f \circ f)(x) = (f \circ f \circ f)(y)$. This information, together with the information you've been given, suffices to derive the fact that $x=y$, and hence that $f$ is injective.
You can compute $f(0)$ using injectivity. Notice that $f(f(f(0)))=f(f(0))+0 = f(f(0))$. Apply the fact that $f$ is injective twice, and you'll find the value of $f(0)$ pretty quickly.