Further if $$f(f(x)) = g(x)$$ and we know that $g(-x)=-g(x)$ , i.e it's an odd function can we determine whether $f$ will be odd or not ? Coming back to the original question, I first determined that $f($ is an injective/one-one function by, $$ f(x) = f(y) $$ $$ \implies f(f(x)) = f(f(y)) $$ $$ \implies x^3 = y^3 $$ $$ \implies x=y$$ $ \therefore f$ is an injective/one-one function. And since $f$ is injective it can't be an even function (except $f\equiv0$)
I couldn't further prove that $f$ will be odd, but what I noticed is; assuming that $f$ is an odd function, we find $$ f(f(-x)) = f(-(f(x)) = -f(f(x)),$$ i.e., $g$ (in this case $g(x) = x^3$) is an odd function which is true. But it doesn't really prove that $f$ has to be an odd function.
Can $f$ be determined to be an odd function or is it impossible to prove, when $f(f(x)) = x^3$ or in the more general case of $f(f(x)) = g(x)$