The problem asks us to find the domain of the function $f(x, y) = \frac{y}{x} + xy$ and evaluate the function at several points. Specifically, we need to find $f(1, 2)$, $f(\frac{1}{4}, 4)$, $f(4, \frac{1}{4})$, $f(a, a)$, $f(\frac{1}{x}, x^2)$, and $f(0, 0)$.

1. Problem Description

The problem asks us to find the domain of the function $f(x, y) = \frac{y}{x} + xy$ and evaluate the function at several points. Specifically, we need to find $f(1, 2)$, $f(\frac{1}{4}, 4)$, $f(4, \frac{1}{4})$, $f(a, a)$, $f(\frac{1}{x}, x^2)$, and $f(0, 0)$.

2. Solution Steps

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

The only restriction is that the denominator $x$ cannot be zero. Therefore, the domain is all $(x, y)$ such that $x \neq 0$.

Now, we evaluate the function at the given points:

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

(b) $f(\frac{1}{4}, 4) = \frac{4}{\frac{1}{4}} + (\frac{1}{4})(4) = 4 \cdot 4 + 1 = 16 + 1 = 17$.

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

(d) $f(a, a) = \frac{a}{a} + (a)(a) = 1 + a^2$ (assuming $a \neq 0$). If $a = 0$, then $f(0,0)$ is undefined.

(e) $f(\frac{1}{x}, x^2) = \frac{x^2}{\frac{1}{x}} + (\frac{1}{x})(x^2) = x^2 \cdot x + x = x^3 + x$. The domain of this expression is $x \neq 0$.

(f) $f(0, 0)$ is undefined because $x$ cannot be zero.

3. Final Answer

(a) $f(1, 2) = 4$

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

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

(d) $f(a, a) = 1 + a^2$ (for $a \neq 0$)

(e) $f(\frac{1}{x}, x^2) = x^3 + x$ (for $x \neq 0$)

(f) $f(0, 0)$ is undefined.