Answer
$(-\infty, -3) \cup (-3, 3) \cup (3,\infty)$
Work Step by Step
Sharp turn at $x=-3$ and $x=3$.
This means that $\lim\limits_{x \to -3^-} f(x) ≠ \lim\limits_{x \to -3^+} f(x)$ and $\lim\limits_{x \to 3^-} f(x) ≠ \lim\limits_{x \to 3^+} f(x)$. Hence it is not differentiable at $x=-3$ and $x=3$
