We wish to prove that $f:\mathbb{R}\setminus \{0\} \to \mathbb{R}, f(x)=\frac{1}{x^2}$ is continuous by definition.
Scratch work:
$$|f(x)-f(a)| =|\frac{1}{x^2} -\frac{1}{a^2}|=|\frac{x^2-a^2}{a^2x^2}|=|x-a|\frac{|x+a|}{a^2x^2}$$
We know that $|x-a|<\delta$ therefore we have that $|a|-\delta <x<|a|+\delta.$
We need to ensure that $a^2x^2 \neq 0$. We know $a \neq 0$ by definition of the functions domain, so we only care that $x \neq 0$.
From the above inequality, we need to have that $|a|-\delta >0$, and this is only possible if $|a|>\delta$. Let us choose $\delta=\frac{|a|}{2}$.
With $\delta$ chosen we see that
$$|x-a|<\frac{|a|}{2} \text{ and that } |x|<\frac{3|a|}{2} $$
Consider $|x+a| \le |x|+|a|$ by the triangle inequality. Then $|x|+|a|<\frac{5|a|}{2}$.
$$\text{Note also that }\frac{1}{x}<\frac{2}{|a|} $$
Here, $$|x-a| \frac{|x+a|}{a^2x^2}=|x-a||x+a|\frac{1}{a^2}\frac{4}{a^2} <|x-a| \frac{5|a|}{2}\frac{1}{a^2}\frac{4}{a^2}=|x-a|\frac{10|a|}{|a|^4}<\delta\frac{10}{|a|^3}$$ We now want to have that $$\delta\frac{10}{|a|^3}<\varepsilon $$ So we choose $$\delta=\min\{\frac{10 \varepsilon}{|a|^3},\frac{|a|}{2}\}$$.
I won't include details of the full proof as it basically just depends on choosing the right $\delta$. Can anyone verify if my steps are correct?
An alternative approach.
$|\frac{1}{x^2}-\frac{1}{a^2}|=|\frac{|a-x||a+x|}{a^2x^2}|\ge\epsilon\implies |x-a|\ge \frac{\epsilon a^2 x^2}{|x+a|}$.
Since $f(x)=f(-x)$, WLOG $a>0$.
Let $-a/2<x-a<a/2$.
Minimize the numerator and maximize the denominator.
$\frac{a^2x^2}{|x+a|}>\frac{a^2(a-a/2)^2}{|a+a/2|}=\frac{a^3}{6}$
So let $\delta=$ min $(a/2,6\epsilon/a^3)$.
This probably suffices since the first statement is the contrapositive of the definition of continuity
We have $ |\frac{1}{x^2}-\frac{1}{\epsilon^2}|\ge \epsilon \implies |x-a|\ge \frac{\epsilon a^2x^2}{|x+a|}>a^3\epsilon/6, $ so by contraposition $|x-a|<a^3\epsilon/6\implies |1/x^2-1/a^2|<\epsilon$, the formal definition of continuity at a point.