Prove by induction that a recurrence has solution $T(n)=\Theta(n^2 \log_{3}n)$
W.l.o.g. suppose $n = 3^k$: The expand the second part of the function:
$$
T(n) = 9 T(\frac{n}{3}) + n^2  = 9 (9T(\frac{n}{3^2}) + (\frac{n}{3})^2) + n^2 = 9^2 T(\frac{n}{3^2}) +‌ 9(\frac{n}{3})^2 + n^2 = 
$$
$$
9^3 T(\frac{n}{3^3}) +‌ 9^2 (\frac{n}{3^2})^2 + 9(\frac{n}{3})^2 + n^2
$$
You can continue it by induction to obtain the following:
$$
T(n) = 9^k T(\frac{n}{3^k}) + 9^{k-1}(\frac{n}{3^{k-1}})^2 + \cdots + 9(\frac{n}{3})^2 + n^2
$$
As $n = 3^k$, $T(\frac{n}{3^k}) = T(1) = 1$. Hence,
$$
T(n) = \sum_{i=0}^k9^i(\frac{n}{3^i})^2 = \sum_{i=0}^k9^i\frac{n^2}{9^i} = \sum_{i=0}^kn^2 = n^2 \times k 
$$
We know that $k = \log_3(n)$. Therefore:
$$
T(n) = n^2 \log_3(n).
$$
and we can say $T(n) = \Theta(n^2 \log(n))$ asymptotically.