If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an alternative what I did is
$y = \frac{x^2 + 2x + 4}{x^2 - 2x + 4}$
Then$ \frac{1+y}{1-y} =\frac{x^2+4}{2x} $
Therefore$ \frac{1+y}{1-y} = {x/2+2/x}$
If we go further putting rhs range to lhs the answer gets altered . Is this method wrong if so why if not where did the process go wrong.

Notice that $f(-x) = \frac{1}{f(x)}$, so the analysis we do for one side will apply to the other (aka take $x\geq0$ WLOG).
We have
$$\frac{x^2-2x+4}{x^2+2x+4} \leq 1$$
with equality achieved at least at $x=0$. Then rewriting the function we have
$$\frac{(x-2)^2+2x}{(x-2)^2+6x} = 1 - \frac{4x}{(x-2)^2+6x} \geq 1 - \frac{4x}{6x} = \frac{1}{3}$$
and in fact this minimum is reached at $x=2$. Therefore
$$1 \geq f(x) \geq \frac{1}{3}$$
on $x\in[0,\infty)$ which implies
$$3 \geq f(x) \geq \frac{1}{3}$$
on $x\in\Bbb{R}$, with max and min achieved at least at $x=\pm2$
