jmc

Since we know that the terms inside any square root need to be greater than or equal to zero, both $x^2-16\ge 0$ and $\sqrt{x^2-16}-3\ge 0$ must hold. Since the first inequality factors as $(x+4)(x-4)\ge 0$, the values of $x$ such that $x^2-16\ge 0$ is $x\le -4$ or $x\ge 4$. Next, we tackle the second inequality: $$\begin{array}{rl}\sqrt{x^2-16}-3& \ge 0\\ \iff \sqrt{x^2-16}& \ge 3\\ \iff x^2-16& \ge 9\\ \iff x^2-25& \ge 0\\ \iff (x+5)(x-5)& \ge 0\end{array}$$This tells us that the domain of $\sqrt{\sqrt{x^2-16}-3}$ is $x\le -5$ or $x\ge 5$. Since this is a subset of the domain we found for the first inequality, these values of $x$ also satisfy $x^2-16\ge 0$. Therefore, the domain of $f(x)$ is $x\in \boxed{(-\infty ,-5]\cup [5,\infty )}$