We are asked to find the partial derivatives of the function $f(x, y) = (4x - y^2)^{3/2}$.

1. Problem Description

We are asked to find the partial derivatives of the function $f(x, y) = (4x - y^2)^{3/2}$.

2. Solution Steps

First, let's find the partial derivative with respect to $x$, denoted as $f_x$.

$f_x = \frac{\partial}{\partial x} (4x - y^2)^{3/2}$

Using the chain rule:

$f_x = \frac{3}{2}(4x - y^2)^{3/2 - 1} \cdot \frac{\partial}{\partial x} (4x - y^2)$

$f_x = \frac{3}{2}(4x - y^2)^{1/2} \cdot 4$

$f_x = 6(4x - y^2)^{1/2}$

Next, let's find the partial derivative with respect to $y$, denoted as $f_y$.

$f_y = \frac{\partial}{\partial y} (4x - y^2)^{3/2}$

Using the chain rule:

$f_y = \frac{3}{2}(4x - y^2)^{3/2 - 1} \cdot \frac{\partial}{\partial y} (4x - y^2)$

$f_y = \frac{3}{2}(4x - y^2)^{1/2} \cdot (-2y)$

$f_y = -3y(4x - y^2)^{1/2}$

3. Final Answer

$f_x = 6(4x - y^2)^{1/2}$

$f_y = -3y(4x - y^2)^{1/2}$