We need to find the first-order partial derivatives for the function $f(x, y) = \frac{x^2 - y^2}{xy}$. That means we need to find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

1. Problem Description

We need to find the first-order partial derivatives for the function $f(x, y) = \frac{x^2 - y^2}{xy}$. That means we need to find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

2. Solution Steps

Let $f(x, y) = \frac{x^2 - y^2}{xy}$.

We can rewrite this as $f(x, y) = \frac{x^2}{xy} - \frac{y^2}{xy} = \frac{x}{y} - \frac{y}{x}$.

First, we find the partial derivative with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\frac{x}{y} - \frac{y}{x})$

$\frac{\partial f}{\partial x} = \frac{1}{y} - y \cdot \frac{-1}{x^2} = \frac{1}{y} + \frac{y}{x^2}$.

Next, we find the partial derivative with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(\frac{x}{y} - \frac{y}{x})$

$\frac{\partial f}{\partial y} = x \cdot \frac{-1}{y^2} - \frac{1}{x} = -\frac{x}{y^2} - \frac{1}{x}$.

3. Final Answer

$\frac{\partial f}{\partial x} = \frac{1}{y} + \frac{y}{x^2}$

$\frac{\partial f}{\partial y} = -\frac{x}{y^2} - \frac{1}{x}$