$$T(n) \in O(n \log n)$$
is defined as
$$(\exists C>0, n_0>0)(\forall n > n_o)\, T(n) \le C \cdot n \log n$$
Given that $T(n) = 3\,T\left(\frac n4\right) + n \log n$, we need to find $C$ and $n_0$ to satisfy the definition. Let's proceed inductively:
$$T(n) \le C \cdot n \log n$$
$$3\,T\left(\frac n4\right) + n \log n \le C \cdot n \log n$$
Now we see that we need to borrow the inductive hypothesis $T\left(\frac n4\right) \le C \cdot \frac n4 \log \frac n4$. Thus the above statement would be implied by:
$$3\,\left(C \cdot \frac n4 \log \frac n4\right) + n \log n \le C \cdot n \log n$$
So now if we can find a positive $C$ that makes the above statement true for sufficiently large positive $n$, then we have satisfied the desired definition. Move things around a bit:
$$3~C~n~\log n - 3~C~n~\log 4 + 4~n~\log n \le 4~C~n~\log n$$
$$4 \le \frac{3~C~\log 4} {\log n } + C$$
$$\frac{4 \log ~ n}{\log n + 3~\log(4)} \le C$$
We can see that the left hand side will never be as large as $4$ (let $n = 4^z$ and simplify to make it more clear), so $C \ge 4$ satisfies the inequality for sufficiently large $n$.