The problem asks for the natural domain of the function $f(x, y) = \frac{y}{x} + xy$, and to evaluate $f(1, 2)$, $f(a, a)$, $f(\frac{1}{x}, x^2)$, $f(4, \frac{1}{4})$, $f(4, 4)$, and $f(0, 0)$.

1. Problem Description

The problem asks for the natural domain of the function $f(x, y) = \frac{y}{x} + xy$, and to evaluate $f(1, 2)$, $f(a, a)$, $f(\frac{1}{x}, x^2)$, $f(4, \frac{1}{4})$, $f(4, 4)$, and $f(0, 0)$.

2. Solution Steps

First, let's determine the natural domain of the function $f(x, y) = \frac{y}{x} + xy$.

Since the term $\frac{y}{x}$ is present, $x$ cannot be equal to

0. There are no other restrictions on $x$ or $y$.

Thus, the domain is $\{(x, y) \in \mathbb{R}^2 : x \neq 0\}$.

Now, we will evaluate the function at the given points.

(a) $f(1, 2) = \frac{2}{1} + (1)(2) = 2 + 2 = 4$

(b) $f(\frac{1}{x}, x^2) = \frac{x^2}{\frac{1}{x}} + (\frac{1}{x})(x^2) = x^3 + x$, provided $x \neq 0$.

(c) $f(a, a) = \frac{a}{a} + (a)(a) = 1 + a^2$, provided $a \neq 0$.

(d) $f(\frac{1}{4}, 4) = \frac{4}{\frac{1}{4}} + (\frac{1}{4})(4) = 16 + 1 = 17$

(e) $f(4, 4) = \frac{4}{4} + (4)(4) = 1 + 16 = 17$

(f) $f(0, 0)$ is undefined, since $x$ cannot be

0.

3. Final Answer

The natural domain of the function is $\{(x, y) \in \mathbb{R}^2 : x \neq 0\}$.

$f(1, 2) = 4$

$f(\frac{1}{x}, x^2) = x^3 + x$

$f(a, a) = 1 + a^2$

$f(\frac{1}{4}, 4) = 17$

$f(4, 4) = 17$

$f(0, 0)$ is undefined.