How to solve $T(n)=4T(\sqrt{n}/3)+(\log n)^2$ with the master theorem?
The division by 3 makes the task a little non-trivial, but you can still figure out how to proceed in the manner proposed by the answer suggested by Nathaniel in the comments.
Let $n = 3^{2^k-2}$ and you can modify the recurrence equation as
$$T(n) = 4T\left(\frac{\sqrt{n}}{3}\right) + (\log_3{2})^2.(\log_3n)^2$$
However, this modification will only bring a change of constant in your final solution. So we can safely assume that the original equation had the base of the logarithm as 3.
You have that $k = \log_2(\log_3n+2)$, and you can substitute this at the end to find your answer.