You need to use Case 1 of the master theorem, which says that if $ T(n)=aT(n/b)+f(n),\qquad a\ge 1, \ b>1, \qquad (1)$ and $f(n)=O(n^{(\log_b a)-\epsilon}), \qquad \epsilon>0,\qquad (2)$ then $T(n)=\Theta(n^{\log_b a}). \qquad (3)$ (Also, some assumptions on the way to round $n/b$ and the positivity of $T$ and $f$ are necessary.)
In this case, $a=3$ and $b=2$, so $\log_b a =\log_2 3 > 1$. Since $f(n)=n\log n$, we have $f(n)=O(n^{1+\delta})$ for any $\delta>0$. Now take $\delta=\epsilon=((\log_b a)-1)/2$ to find that (2) holds. Then. by (3), $T(n)=\Theta(n^{\log_2 3}).$
To find the order of $T(n)$ without using the master theorem, you can do the following:
Find a constant $\epsilon>0$ such that, if $2^m\le n\le 2^{m+1}$ and $m\ge 0$, then $T(n)\ge \epsilon 3^m$. Then prove by induction that this is true for all $m\ge 0$. This proves that $T(n)=\Omega(n^{\log_2 3}).$
Find constants $K>0$, $L>0$, and $M\ge 0$ such that, if $2^m\le n\le 2^{m+1}$ and $m\ge M$, then $T(n)\le K 3^m - L m 2^m $. Then prove by induction that this is true for all $m\ge M$. This proves that $T(n)=O(n^{\log_2 3}).$
Together, this proves that $T(n)=\Theta(n^{\log_2 3}).$