Finding the range of $\frac{x^2}{x^2-9}$ I am a student who is studying about functions (only basic ones) and was practicing till I found this -
Finding the range of 
$$ \frac{x^2}{x^2-9} $$ 
Since I have only learnt the basics , I can only play around with the numbers and not use limits (which was what I found online). 
I was Told to try this method -  to try to assign $y=\frac{x^2}{x^2-9}$ 
This means expressing $x$ in terms of $y$ 
$x = \sqrt { \frac{9y}{y-1} } $
Now I then go on to find the range of this function - 
range =$ y \leq 0  $ or $ y>1$ 
This is simple to do. 
But what I'm confused with is this - 
Since Range is all the possible 'y' values obtainable from the domain, then this confuses me because I express x in terms of y. Finding the x value feels like finding the 'domain' for me . 
I believe I'm having a conceptual problem and I don't understand what does expressing x in terms of y do to help me find the range. does this mean that I can express all other simple functions into x in terms of y to find the range too ? 
Thanks !! 
Note : I'm being taught on How to read the domain And range off a function. So I can't use the graphical method . 

Since you already have $$x = \sqrt { \frac{9y}{y-1} }$$
you can see that $$\sqrt { \frac{9y}{y-1} }$$ exists iff $$\frac{y}{y-1}\ge0~~~ \&~~ y\neq 1 \Longleftrightarrow y(y-1)\ge 0 ~~~ \&~~ y\neq 1 \Longleftrightarrow y\in\Bbb R\setminus(0,1]$$
That is your range is $$\color{blue}{\Bbb R\setminus(0,1] = (-\infty, 0]\cup(1,\infty)}$$
