Lets give some meaning to the whole process.

Domain of the function $\displaystyle g(x)=\log_4\left(\frac{x+3}{x}\right)$ is clearly dictated by the logarithm and the fraction *beneath* it. In order to calculate the function value one must be able to calculate the value of the logarithm. That can be done **a)** if $\displaystyle x\neq 0$ and **b)** if and only if the argument of the logarithm belongs to the domain of the logarithm. That gives us the inequality:

$\displaystyle \frac{x+3}{x}>0$

One way to think of this inequality is as follows:

the quotient of the expressions (x+3) and x must be positive and that will be the case only if both expressions have the same sign, meaning they both simultaneously must be either positive or negative. Hence:

$\displaystyle (x+3> 0\quad and \quad x>0)\quad or \quad (x+3< 0 \quad and \quad x<0)$

$\displaystyle (x> -3\quad and \quad x>0)\quad or \quad (x< -3 \quad and \quad x<0)$

$\displaystyle ( x>0 )\quad or \quad (x< -3) $

Since Mathematics is a language one should translate/substitute every 'and' with an intersection, and every 'or' with a union. Hence,

$\displaystyle \langle0,+\infty \rangle \cup \langle -\infty, -3\rangle $ or reordering that $\displaystyle \langle -\infty, -3\rangle \cup \langle0,+\infty \rangle $