Answer
$\delta = \sqrt{9+\epsilon}-3$ is the largest possible choice of $\delta$
Work Step by Step
Let $\delta = \sqrt{9+\epsilon}-3$
Suppose that $~~x = 3+\delta$
Then:
$x^2-9$
$ = (3+\delta)^2-9$
$ = (3+\sqrt{9+\epsilon}-3)^2-9$
$ = (\sqrt{9+\epsilon})^2-9$
$ = 9+\epsilon-9$
$ = \epsilon$
Note that $x^2$ is an increasing function for all $x \gt 0$
If $3 \lt x \lt 3+\delta$, then $0 \lt x^2-9 \lt \epsilon$
Thus $\delta = \sqrt{9+\epsilon}-3$ is the largest possible choice of $\delta$ when we consider the values of $x$ such that $3 \lt x \lt 3+\delta$
Note that because the slope of the $x^2$ continues increasing more steeply as $x$ becomes more positive, then $0 \lt \vert (3-\delta)^2-9 \vert \lt \vert (3+\delta)^2 - 9 \vert = \epsilon$
Thus $\delta = \sqrt{9+\epsilon}-3$ is the largest possible choice of $\delta$
If $0 \lt \vert x-3 \vert \lt \delta,~~$ then $~~0 \lt \vert x^2-9 \vert \lt \epsilon$
