Upper bound $T(n) = 9T(\sqrt[3]{n}) + O(1)$
You can solve this using the master theorem, whose proof uses a recursion tree argument.

Define $S(n) = T(2^n)$. The main observation is that $\sqrt[3]{2^n} = 2^{n/3}$, and so
$$ S(n) = 9S(n/3) + O(1). $$
The solution is $S(n) = \Theta(n^2)$, and so
$$ T(n) = S(\log n) = \Theta(\log^2 n).$$

You can also open the recurrence directly, obtaining
$$
T(n) \propto 1 + 9n^{1/3} + 9^2 n^{1/3^2} + 9^3 n^{1/3^3} + \cdots.
$$
The general term is $9^i n^{1/3^i}$. The recurrence will terminate when $n^{1/3^i} \approx 1$, which happens when $3^i \approx \log n$, that is, when $i \approx \log_3 \log n$. For this value of $i$, $$ 9^i = 3^{2\log_3 \log n} = \log^2 n. $$
This is the value of the bottom level of the tree. The contribution of the other levels is negligible in comparison, which is something you have to prove by calculation.