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 . 

The domain is the set of $x$ values such that the function is defined. Here the expression exists for any $x\notin \{-3,3\}$.
The range is the set of $y$ values for which the equation $y=f(x)$ has solutions.
Indeed, 
$$y=\frac{x^2}{x^2-9}$$ can be written 
$$x^2=\frac{9y}{y-1}$$ and the RHS is only non-negative and defined for $y\notin(0,1]$.
Remember: the domain is for $x$, the range for $y$.
Alternatively, you can study the variations of $f$. Canceling the derivative,
$$f'(x)=-\frac{18x}{(x^2-9)^2}$$ and noting the two vertical asymptotes and the horizontal one, the table of variations is
$$\begin{matrix}x&-\infty&&&-3&&&0&&&3&&&\infty\\\hline f(x)&1&\nearrow&\infty&|&-\infty&\nearrow&0&\searrow&-\infty&|&\infty&\searrow&1\end{matrix}$$
Then all values are reached, except for $(0,1]$.
