Understanding how this derivative was taken 
I am pouring water into a conical cup 8cm tall and 6cm across the top.
  If the volume of the cup at time t is $V(t)$, how fast is the water
  level ($h$) rising in terms of $V'(t)$?

The solution in the book is:

Take the water volume, given by
$$\frac{1}{3}\pi(\frac{3}{8})^2h^3$$
Then differentiate with respect to $t$:
$$V' = h'\pi(\frac{3}{8})^2h^2$$
Which gives
$$h' = \frac{64V'}{9{\pi}h^2}$$

I did not understand how this differentiation happened, when there was no $t$ in the formula to differentiate! If you differentiate with respect to $h$, though, you get something similar:
$$\dfrac{9{\pi}h^2}{64}$$
But I'm not sure how $V'$ fits into this.
Thanks in advance!

To answer in words, making GoodDeeds point more clear.
Notice how if you differentiate the whole term with respect to t
$$
V = \frac{1}{3}{\pi}(\frac{3}{8})^2h^3
$$
use chain rule on the h such that the expression becomes
$$
h'*constants *3h^2
$$
where h' is the derivative of h with.
All thats left is to deal with the constants and rearrange.
