First at all,
Let $\lim x→0 \frac{x-1}{x^2-1}=1$ then there exists $\delta>0$ such that if $0<|x|<\delta$ then $|f(x)-1|< \varepsilon $
Proof:
$\bigg|\frac{x-1}{x^2-1}-1\bigg|=\bigg|\frac{x}{x}\big(\frac{1-1/x}{x-1/x} \big)-1\bigg|=\bigg|\frac{1}{x-1/x}-\frac{1/x}{x-1/x}-1 \bigg|=\bigg|\frac{1}{x-1/x}-x^2+1-1 \bigg|=\bigg|\frac{1}{x-1/x}-x^2 \bigg|$
I use the triangular inequality,
$\bigg|\frac{1}{x-1/x}-x^2 \bigg|\leq\bigg|\frac{1}{x-1/x}\bigg|+\bigg|-x^2 \bigg|=\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x^2 \bigg|=\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x \bigg|^2$
Let say $\frac{1}{x-1/x}<1$. If that's the case then:
$1<x-\frac{1}{x} \Longrightarrow 1+\frac{1}{x}<x \Longrightarrow \frac{x+1}{x}<x \Longrightarrow x+1<x^2 \Longrightarrow 0<x^2-x-1$
Using the quadratic formula, I know that inequality is true for $x:(-1.6 ; 0.6)$ approximately, which is good knowing that I'm interested in those $x$ near $0$. Then,
$\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x \bigg|^2\leq1+|x|^2<\varepsilon$
I take $\delta=\sqrt{\varepsilon -1}$ and QED, which is weird because $\varepsilon$ should be bigger than $1$, and isn't that a bit big?
I'm sure there's a simpler way to do this. But I'd like to know if it's ok, or where did I go wrong...
In your first line you replaced $\frac {1/x} {x-\frac 1 x}$ by $x^{2}-1$ which is not correct. Let $0 <\epsilon <1$ Then $|f(x)-1|=|\frac 1 {x+1} -1|=\frac {|x|} {|x+1|} \leq \frac {|x|} {1-|x|}<\epsilon$ if $|x| <\frac {\epsilon} {1+\epsilon}$. Take $\delta =\frac {\epsilon} {1+\epsilon}$.