$T(n)=n\displaystyle \cdot T\left(\frac{n}{2}\right)+n^{\log_{2}n}$.
$f(n) = n^{\log_{2}n}$
Number of leaves = $n^{\log_{a}b} = n^{\log_{2}n}$
CASE 2 (All level same)
$f(n) = \Theta(n^{\log_{b}a} {\log^{k}n}) $
$f(n) = \Theta(n^{\log_{2}n} {\log^{0}n}), $ because $b = 2$, $a = n$, $k = 0$
Is $T(n) = \Theta(n^{\log_{2}n} {\log_{2}n)} $ correct?
This recurrence cannot be solved with the master theorem, since the master theorem only applies to recurrences of the form $T(n) = aT(n/b) + f(n)$ where $a\geq1$ and $b>1$ are constants. Your recurrence is not of this form, so the theorem doesn't apply.
Let $T(n) = n⋅T(n/2) + n^{log_2n}$, and let $T(n) = (1 + C(n))·n^{log_2n}$.
Then $T(n) = n⋅(1 + C(n/2))·(n/2)^{log_2n - 1} + n^{log_2n}$
$T(n) = 2⋅(1 + C(n/2))·(n/2)^{log_2n} + n^{log_2n}$
$T(n) = (1 + C(n/2)) / 2^{log_2n - 1} ·n^{log_2n} + n^{log_2n}$
$T(n) = (1 + (1 + C(n/2)) / 2^{log_2n-1}) · n^{log_2n}$
$C(n) = (1 + C(n/2)) / 2^{log_2n-1} = (1 + C(n/2)) · (2/n)$
Clearly C(n) -> 0, therefore $T(n) = \Theta(n^{log_2n})$. Actually, the limit of $T(n) / n^{log_2n}$ is 1.
$n^{log_2n}$ grows so fast that n·T(n/2) cannot keep up with it. For T(n) to grow faster than $O(n^{log_2n})$, n·T(n/2) would have to grow faster, and it doesn't.
