#### Answer

a) $y=f(x)$ is differentiable on $[-3,0)\cup(0,3]$.
b) There are no domain points in $[-3,3]$ where $y=f(x)$ is continuous but not differentiable.
c) $y=f(x)$ is neither continuous nor differentiable at $x=0$.

#### Work Step by Step

*Some things to remember about differentiability:
- If $f(x)$ is differentiable at $x=c$, then $f(x)$ is continuous at $x=c$. (Theorem 1)
- $f(x)$ is not differentiable at $x=c$ if the secant lines passing $x=c$ fail to take up a limiting position or can only take up a vertical tangent. In other words, we can look at differentiability as the ability to draw a tangent line at a point, or the smoothness of the graph.
a) In this exercise, we have a discontinuous curve $y=f(x)$ on the closed interval $[-3,3]$. $f(x)$ has a smooth continuous curve on $[-3,0)$, then the graph jumps to $0$ at $x=0$, and then it jumps again to continue another smooth continuous curve on $(0,3]$.
We see that on $[-3,0)\cup(0,3]$, the graph is all continuous and smooth; there are no corners or cusps or any points having vertical tangents.
So $y=f(x)$ is differentiable on $[-3,0)\cup(0,3]$.
For $x=0$, $f(x)$ is not continuous here, meaning that $f(x)$ is also not differentiable at $x=0$.
b) There are no domain points in $[-3,3]$ where $y=f(x)$ is continuous but not differentiable. Where $f(x)$ is continuous, it is differentiable.
c) $y=f(x)$ is neither continuous nor differentiable at $x=0$.