Solving $T(n)=3T(\lfloor \frac{n}{3}\rfloor) +2n\log n$
Ignoring floors (see, e.g., the paragraph &quot;Ignoring Floors and Ceilings Is Okay, Honest&quot; in Chapter 1 of Algorithms for a discussion), your recurrence is of the form
$$
T(n) = aT(n/b) + f(n),
$$
with $a=b=3$ and $f(n) = 2 n \log n$.
You can then immediately apply the master theorem since, for $k=1$, $f(n) \in \Theta(n^{\log_b a} \log^k n) = \Theta(n \log n)$.
The solution to the recurrence is then:
$$
T(n) \in \Theta(n^{\log_b a}  \log^{k+1} n) = \Theta(n \log^2 n).
$$