How to solve this algebra problem

I have what I consider a difficult algebra question. It turns into an inception of fractions, just fractions within fractions within fractions and I can't keep straight what to do. So could use a little guidance.

The question is (4z^(3/4))^2 over (2z^(-1/3))^-2 . For ease of reading, Top 4z is to the 3/4 power and bottom 2z is to the -1/3 power. Whole part on top is squared, and whole part on bottom is to the -2 power.

Thanks for any guidance

(4z^(3/4))^2 over (2z^(-1/3))^-2 = [(4z^(3/4))^2][(2z^(-1/3))^2] =

[16z^(3/2)][4z^(-2/3)] = 64z^(5/6)

((4z)^(3/4))^2 / ((2z)^(-1/3))^-2
=> ((4z)^(3/4))^2 X ((2z)^(-1/3))^2
=> (4z)^(3/2) X (2z)^(-2/3)
=> 4^(3/2) X z^(3/2) X 2^(-2/3) X z^(-2/3)
but 4^(3/2) can be written as 2^(6/2),
=> 2^(6/2) X z^(3/2) X 2^(-2/3) X z^(-2/3)
law of indices=> 2^(6/2-2/3) X z^(3/2-2/3)
Ans=>2^(7/3) X z^(5/6)
So simple?

16 z^(3/2) x 4
------------------
z^(2/3)

64 z^(5/6)