My reasoning is: suppose both limits exist, $\lim_{x\to 0} f(x) = l$ and $\lim_{x\to 0} f(x^3) = m$.
For the second limit, by the limit definition we have:
$$\forall\epsilon>0,\ \exists\delta,\ \forall x,\ 0<|x-0|<\delta \implies |f(x^3)-m|<\epsilon.$$
Next I claim that $0<|x|<\delta \iff 0<|x^3|<\delta^3=\delta_1$ (omitting the proof for brevity), so that $$\forall\epsilon>0,\ \exists\delta_1,\ \forall x,\ 0<|x^3|<\delta_1 \implies |f(x^3)-m|<\epsilon.$$
Now, by the limit definition, we have $\lim_{x^3\to 0} f(x^3) = m$, and by the theorem that a function cannot approach two different limits at some $a$, I conclude that $l=m$.
Changing $3$ to $n$ gives proof for any $n\in N$ (with the help of $0<|x|<\delta \iff 0<|x^n|<\delta^n, n\in N$).
I that reasoning correct?