Epsilon-delta for x squared

Scratch work. We have $|x^2-9|=|x-3|\,|x+3|$. The factor $|x-3|$ is the one we control. The factor $|x+3|$ is not constant, but it is bounded if $x$ stays near $3$.

If $|x-3|<1$, then $2<x<4$, hence $5<x+3<7$ and $|x+3|<7$. So under this restriction,
$$|x^2-9|<7\,|x-3|.$$
We want this to be at most $\varepsilon$, so we ask for $|x-3|<\frac\varepsilon7$ as well.

Proof. Let $\varepsilon>0$ and choose $\delta=\min\!\left(1,\dfrac\varepsilon7\right)$. If $|x-3|<\delta$, then $|x-3|<1$, so $|x+3|<7$, and
$$|x^2-9|=|x-3|\,|x+3|<\delta\cdot7\le\frac\varepsilon7\cdot7=\varepsilon.$$

Remark. Choosing $\delta$ as a minimum of two requirements is a standard technique. Here the first one ($\delta\le1$) controls a factor, the second one ($\delta\le\varepsilon/7$) makes the product small.