Question: If $h(x)=f(g(f(x)))$ as a function $\mathbb R \rightarrow \mathbb R$ is bijective, what do we know about $f,g$, which are also functions $\mathbb R \rightarrow \mathbb R$? Is my proof correct?
My attempt: $f$ must be injective. If not, there exist $a,b$ such that $f(a)=f(b)$, but then $f(g(f(a)))=f(g(f(b)))$, which contradicts the fact that $h$ is injective.
$f$ must be surjective. If there is a value that is not taken by $f$, then it is certainly not taken by $h(x)=f(g(f(x)))$, contradicting that it is bijective.
Suppose there exist $a,b$ such that $g(a)=g(b)$. Because $f$ is surjective, $a,b$ are in the range of $f$, and thus there exist $c,d$ such that $g(f(c))=g(f(d))$, and thus $f(g(f(c)))=f(g(f(d)))$, which contradicts the fact that $h$ is injective. Therefore $g$ is injective.
Since $f$ is bijective, there is an unique $x$ such that $f(x)=p$. If $x$ is not in the range of $g$, then $p$ is not in the range of $h(x)=f(g(f(x)))$, contradicting that it is surjective. Therefore all real numbers are in the range of $g$, thus $g$ is surjective.
Therefore both $f$ and $g$ are bijective.
Is this proof correct?
Your proof is fine. It's also worth noting that proving $g$ is bijective could be done by considering the inverse of $f$ (there is an inverse because $f$ is bijective). Then, $$h = f \circ g \circ f \implies f^{-1} \circ h \circ f^{-1} = f^{-1} \circ f \circ g \circ f \circ f^{-1} = g,$$ so $g$ is the composition of three bijections, and hence is bijective.