Not very sure if this is correct.
y = (x^2)^(ln x)
ln both sides:
ln y = 2 (ln x) (ln x)
ln y = 2 (ln x)^2
1/y (dy/dx) = 4 (ln x) (1/x)
dy/dx = (y)
(4/x) (ln x)
dy/dx = (x^2)^(ln x) (4/x) (ln x)
dy/dx = [4(x^2)^(ln x) * (ln x)] / x
My algebra not powerful enough to simplify to your answer,
provided my answer is correct. 
Maybe you wait for Mr. Wee or eagle to come help.
Continuing from the underlined equation,
dy/dx = 4 (Inx) ( x^-1) ( x^ (2Inx))
dy/dx = 4 (Inx) (x ^ (2Inx - 1))
dy/dx = 4 (In (x^(x^(2Inx - 1))))
It is still different from the answer of 4(In (x^(2Inx -1)))
though.
