Related Rates Problem involving a cone

A scoop of ice cream is in a waffle cone of radius 3 in. and height of 8in. It begins to melt and drain out of a hole in the bottom at a rate of .4in/min. Find the rate at which the height of the ice cream is changing when the ice cream has reached the point on the cone of radius 1.5in.

Here's what I have so far:

dv/dt = -.4
V= (1/3)π r² h

Using similar triangles I have:
r = 3h/8
h = 8r/3

Replace r with 3h/8
V = (1/3)π (3h/8)² h

Which will give me:
V = (3/64)π h^3

I take the derivative with respect to time [ (3/8)π are constants so I move them to the front]:
V' = (3/64)π 3h² h'

I solve for h':
h' = (64V' / 9π h²)

This is where I'm stuck, I have no idea where to go from here =/

Any help would be much appreciated!

This is good so far.

v' = dV/dt= -.4 and when r= 1.5, or 3/2, h= (8/3)(3/2)= 4

Plug these in to get h' = 64* (-.4) /[ 9pi* 16]

= -1.6/ (9pi)

Or -8/(45pi) in/min

Hoping this helps!