If $f(2^x-3x)=2^{f(x)}-3f(x), \forall x\in\mathbb{R}$, prove that there exists $\xi\in\mathbb{R}$ such that $f(f(\xi))=\xi$ 
Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that
  $f(2^x-3x)=2^{f(x)}-3f(x), \forall x\in\mathbb{R}$.
Prove that there is at least one $\xi\in\mathbb{R}$ such that
  $f(f(\xi))=\xi$.

I have observed that if $g(x)=2^x-3x$ then $f\circ g=g\circ f$, but don't see to move any further... Any hint?

Your observation is spot-on, but yo need to have a look at the properties of $g$:
One can verify (e.g. from a rough sketch of the graph of $x\mapsto g(x)-x$) that the function $g$ has exactly two fixed points $\xi_1,\xi_2$. But if $g(x)=x$ then $g(f(x))=f(g(x))=f(x)$, i.e., $f$ maps a fixed point of $g$ to a fixed point of $g$. If at least one of $\xi_1,\xi_2$ is also a fixed point of $f$, we are done. And if not $f$ must switch these two points, i.e., $f(\xi_1)=\xi_2$ and $f(\xi_2)=\xi_1$ so that $f(f(\xi_1))=\xi_1$.
