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

1. Problem Description

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

2. Solution Steps

We need to divide each term of the polynomial by $-\frac{1}{5}xy^2$.

$\frac{2x^4y^3}{-\frac{1}{5}xy^2} = 2 \cdot (-5) \cdot \frac{x^4}{x} \cdot \frac{y^3}{y^2} = -10x^{4-1}y^{3-2} = -10x^3y$

$\frac{-\frac{1}{5}x^3y^2}{-\frac{1}{5}xy^2} = \frac{1}{5} \cdot 5 \cdot \frac{x^3}{x} \cdot \frac{y^2}{y^2} = x^{3-1} \cdot 1 = x^2$

$\frac{\frac{1}{4}x^2y}{-\frac{1}{5}xy^2} = \frac{1}{4} \cdot (-5) \cdot \frac{x^2}{x} \cdot \frac{y}{y^2} = -\frac{5}{4}x^{2-1}y^{1-2} = -\frac{5}{4}xy^{-1} = -\frac{5x}{4y}$

So, the result of the division is $-10x^3y + x^2 - \frac{5x}{4y}$.

3. Final Answer

The final answer is $-10x^3y + x^2 - \frac{5x}{4y}$.