It's not solvable by Master Theorem and I couldn't find any solution on the web. I am not familiar with generating functions so it would be nice if you can post a solution without them.
I found the top answer here a little helpful https://stackoverflow.com/questions/25717513/solving-the-recurrence-tn-tn-3-t2n-3-n2 Based on this answer I have tried this,
We have that,
T(n) = T(n / 3) + T(2n / 3)
Consider this other recurrence:
S(n) = S(2n / 3) + S(2n / 3) = 2S(2n / 3)
Since T(n) is increasing and T(n) ≤ S(n), any upper bound for S(n) should also be an upper-bound for T(n).
Using the Master Theorem on S(n), we have that a = 2, b = 3/2, and c = 0. Since logb a = log3/2 2 = 1.709511291... < c, the Master Theorem says that this will solve to O(n^1.70). Since S(n) = O(n^1.70), we also know that T(n) = O(n^1.70). Is this method correct?
EDIT: I also found this https://cs.stackexchange.com/questions/22809/recurrence-of-tn-tn-3-t2n-3 but couldn't proceed even with the hint.