Finding the range of given function

Find the range of $f(x) = \frac{2}{x^2 -16}$

I found that the domain is $(-\infty,-4) \cup(-4,4) \cup (4, \infty)$. Then tried to find the range case by case

$1.$case : if $-\infty <x<-4$ , then $16 <x^2 < \infty$ ,

$0 <x^2 -16 < \infty \rightarrow 0 <\frac{x^2 -16}{2} < \infty \rightarrow \frac{1}{\infty} < \frac{2}{x^2 -16} < \frac{1}{0} = 0 < \frac{2}{x^2 -16} < \infty $

$2.$case : if $-4 <x < 4$ , then $0 \leq x^2 <16$,

$-16 \leq x^2 -16 <0 \rightarrow -8 \leq \frac{x^2 -16}{2} <0 \rightarrow -8\geq \frac{2}{x^2 -16} >0$ but this case is impossible because it cannot be $0 <f(x) \leq -8$

$3.$case : $4<x<\infty$ , then $16 <x^2 < \infty \rightarrow 0< x^2 -16 < \infty \rightarrow 0< \frac{x^2 -16}{2}< \infty \rightarrow \frac{1}{\infty} < \frac{2}{x^2 -16} < \frac{1}{0} \rightarrow 0< f(x) <\infty$

So the only range i found is $(0,\infty)$ , but the answer is $$\bigg(-\infty ,\frac{-1}{8}\bigg) \cup \bigg(0,\infty\bigg)$$

There is a solution for general in here : How to properly write a solution to the following inequality?

However i am interested in "what am i missing" ? Is my solution way not correct ?

Solution 1:

Lets say $$y = \frac{2}{x^2 -16}$$

Now , write $y$ in terms of $x$ such that $$x = \sqrt{\frac{2+16y}{y}}$$

we know that $\frac{2+16y}{y} \geq 0$ because of root square and $y \neq0$ because of denominator , so by solving this in equality we found that $$y \in (-\infty , -1/8] \cup (0, \infty )$$