Let $f:\mathbb R\to\mathbb R$ be a continuous function such that $f(f(f(x)))=-8x$. Must we have $f(x)=-2x$? I can prove this if I assume that $f$ is continuously differentiable everywhere, but is that condition necessary? I was inspired to ask this question after I saw the thread a continuous function satisfying $f(f(f(x)))=-x$ other than $f(x)=-x$ .
Some facts I've discovered, which you may use without proof:
$f$ is bijective, decreasing and therefore $f$ is a homeomorphism and differentiable almost everywhere
$f(0)=0$ because if $f(0)=a$, then $f(f(a))=0$, so $-8a=f(f(f(a)))=f(0)=a\implies a=0$.
If $x\neq 0$, then the list $\ldots,f^{-2}(x),f^{-1}(x),x,f(x),f^2(x),\ldots$ has no duplicate members, where $f^n$ is an $n$th functional power of $f$.
If $f$ is differentiable at $0$, then $f'(0)=-2$ because $f'(0)f'(0)f'(0)=-8$.
$x$ and $f(x)$ have opposite signs if $x\neq 0$, because $f(0)=0$ and $f$ is decreasing.
$f\circ f$ restricted to the positive numbers is an increasing function.
If we assume $f$ is continuously differentiable everywhere, then $f(x)=-2x$. Proof: Let $a_n=f^n(x)$ for some $x\neq 0$. Then $f'(a_1)f'(a_2)f'(a_3)=-8$, and $f'(a_2)f'(a_3)f'(a_4)=-8$ which gives us $f'(a_k)=f'(a_{k+3})$ for all integers $k$. Since $a_{3k}=(-8)^kx\to 0$ when $k\to-\infty$, we get $f'(x)=\lim_{k\to-\infty}f'(f^{-3k}(x))=f'(0)=-2$ and therefore $f(x)=-2x$ is the only solution.