Answer
Critical points $x=0,\pm1$. No endpoint.
At $x=0$, $y=0$, local maximum.
At $x=-1$, $y=-3$, local and absolute minimum..
At $x=1$, $y=-3$, local and absolute minimum.
Work Step by Step
Step 1. Given the function $y=x^{2/3}(x^2-4)$, we can find its derivative as $y'=\frac{2}{3}x^{-1/3}(x^2-4)+x^{2/3}(2x)=\frac{2x^2-8}{3\sqrt[3] x}+\frac{6x^2}{3\sqrt[3] x}=\frac{8(x^2-1)}{3\sqrt[3] x}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=0,\pm1$. The domain of the function is $(-\infty,\infty)$ and there is no endpoint.
Step 3. At $x=0$, $y=0$ and we can identify this as a local maximum (use local test points as necessary) as shown in the figure.
Step 4. At $x=-1$, $y=-3$ and we can identify this as a local and absolute minimum (use local test points as necessary) as shown in the figure.
Step 5. At $x=1$, $y=-3$ and we can identify this as a local and absolute minimum (use local test points as necessary) as shown in the figure.