Answer
$g(x)$ is not differentiable at the origin.
Work Step by Step
$g(x)=x^{2/3}$ for $x\ge0$ and $g(x)=x^{1/3}$ for $x\lt0$
- The right-hand derivative of $g(x)$ at the origin:
As $h\to0^+$, $g(h)=h^{2/3}$ and $g(0)=0^{2/3}=0$. Therefore, $$\lim_{h\to0^+}\frac{g(h)-g(0)}{h}=\lim_{h\to0^+}\frac{h^{2/3}-0}{h}=\lim_{h\to0^+}\frac{1}{h^{1/3}}=\infty$$
- The left-hand derivative of $g(x)$ at the origin:
As $h\to0^-$, $g(h)=h^{1/3}$ and $g(0)=0$. Therefore, $$\lim_{h\to0^-}\frac{g(h)-g(0)}{h}=\lim_{h\to0^-}\frac{h^{1/3}-0}{h}=\lim_{h\to0^-}\frac{1}{h^{2/3}}=\infty$$
Since both the left-hand and the right-hand derivative do not exist, we conclude that $g(x)$ is not differentiable at the origin.
