The problem is to divide the polynomial $\frac{1}{3}y^5 - \frac{2}{3}y^3 + \frac{3}{8}a^2b^2$ by $\frac{1}{4}y^2$.

1. Problem Description

The problem is to divide the polynomial $\frac{1}{3}y^5 - \frac{2}{3}y^3 + \frac{3}{8}a^2b^2$ by $\frac{1}{4}y^2$.

2. Solution Steps

First, we write the expression for the division:

$$(\frac{1}{3}y^5 - \frac{2}{3}y^3 + \frac{3}{8}a^2b^2) \div (\frac{1}{4}y^2)$$

Next, we divide each term of the polynomial by $\frac{1}{4}y^2$.

Term 1: $\frac{1}{3}y^5 \div \frac{1}{4}y^2 = \frac{1}{3}y^5 \cdot \frac{4}{1}y^{-2} = \frac{4}{3}y^{5-2} = \frac{4}{3}y^3$

Term 2: $-\frac{2}{3}y^3 \div \frac{1}{4}y^2 = -\frac{2}{3}y^3 \cdot \frac{4}{1}y^{-2} = -\frac{8}{3}y^{3-2} = -\frac{8}{3}y$

Term 3: $\frac{3}{8}a^2b^2 \div \frac{1}{4}y^2 = \frac{3}{8}a^2b^2 \cdot \frac{4}{1}y^{-2} = \frac{12}{8}a^2b^2y^{-2} = \frac{3}{2}a^2b^2y^{-2} = \frac{3a^2b^2}{2y^2}$

Now, we add the results of the divisions of each term:

$$\frac{4}{3}y^3 - \frac{8}{3}y + \frac{3a^2b^2}{2y^2}$$

3. Final Answer

The final answer is $\frac{4}{3}y^3 - \frac{8}{3}y + \frac{3a^2b^2}{2y^2}$.