The problem asks us to find all first partial derivatives of the given functions. We will solve problems 3, 4, 6 and 7.

1. Problem Description

The problem asks us to find all first partial derivatives of the given functions. We will solve problems 3, 4, 6 and

7.

2. Solution Steps

Problem 3: $f(x, y) = \frac{x^2 - y^2}{xy}$

We first compute the partial derivative with respect to $x$, $f_x$.

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

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

$f_x(x, y) = \frac{1}{y} + \frac{y}{x^2}$

$f_x(x, y) = \frac{x^2 + y^2}{x^2y}$

Next, we compute the partial derivative with respect to $y$, $f_y$.

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

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

$f_y(x, y) = -\frac{x}{y^2} - \frac{1}{x}$

$f_y(x, y) = -\frac{x^2 + y^2}{xy^2}$

Problem 4: $f(x, y) = e^x \cos y$

We first compute the partial derivative with respect to $x$, $f_x$.

$f_x(x, y) = \frac{\partial}{\partial x}(e^x \cos y)$

$f_x(x, y) = e^x \cos y$

Next, we compute the partial derivative with respect to $y$, $f_y$.

$f_y(x, y) = \frac{\partial}{\partial y}(e^x \cos y)$

$f_y(x, y) = e^x (-\sin y)$

$f_y(x, y) = -e^x \sin y$

Problem 6: $f(x, y) = (3x^2 + y^2)^{-1/3}$

We first compute the partial derivative with respect to $x$, $f_x$.

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

$f_x(x, y) = -\frac{1}{3} (3x^2 + y^2)^{-4/3} (6x)$

$f_x(x, y) = -2x (3x^2 + y^2)^{-4/3}$

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

Next, we compute the partial derivative with respect to $y$, $f_y$.

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

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

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

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

Problem 7: $f(x, y) = \sqrt{x^2 - y^2}$

We first compute the partial derivative with respect to $x$, $f_x$.

$f(x, y) = (x^2 - y^2)^{1/2}$

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

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

$f_x(x, y) = \frac{x}{\sqrt{x^2 - y^2}}$

Next, we compute the partial derivative with respect to $y$, $f_y$.

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

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

$f_y(x, y) = \frac{-y}{\sqrt{x^2 - y^2}}$

3. Final Answer

Problem 3:

$f_x(x, y) = \frac{x^2 + y^2}{x^2y}$

$f_y(x, y) = -\frac{x^2 + y^2}{xy^2}$

Problem 4:

$f_x(x, y) = e^x \cos y$

$f_y(x, y) = -e^x \sin y$

Problem 6:

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

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

Problem 7:

$f_x(x, y) = \frac{x}{\sqrt{x^2 - y^2}}$

$f_y(x, y) = \frac{-y}{\sqrt{x^2 - y^2}}$