Answer
$y(0)=0$ is a local maximum and $y(2)=-3\sqrt[3] 4$ is a local minimum; there are no absolute extrema.
$x=-1$ is an inflection point.
See graph.
Work Step by Step
Step 1. Given the function $y=x^{2/3}(x-5)$, we have $y'=x^{2/3}+\frac{2}{3}x^{-1/3}(x-5)=\frac{x}{\sqrt[3] x}+\frac{2x-10}{3\sqrt[3] x}=\frac{5(x-2)}{3\sqrt[3] x}=\frac{5}{3}(x-2)x^{-1/3}$ and $y''=\frac{5}{3}x^{-1/3}-\frac{5}{9}(x-2)x^{-4/3}=\frac{5x}{3\sqrt[3] {x^4}}-\frac{5x-10}{9\sqrt[3] {x^4}}=\frac{10(x+1)}{9\sqrt[3] {x^4}}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,2$ as critical points with $y(0)=0, y(2)=-3\sqrt[3] 4$
Step 3. Examine the sign change of $y'$ across the critical points: $..(+)..(0)..(-)..(2)..(+)$; we can identify $y(0)=0$ as a local maximum and $y(2)=-3\sqrt[3] 4$ as a local minimum. The function does not have absolute extrema as the end behaviors are $x\to\pm\infty, y\to\pm\infty$.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=-1,0$ as possible inflection points.
Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(-)..(-1)..(+)..(0)..(+)..$,
Step 6. The function is concave down on $(-\infty,-1)$ and concave up on $(-1,0), (0,\infty)$; this also means that $x=0$ is not an inflection point (no concavity change) and $x=-1$ is an inflection point.
Step 7. See graph.
