Answer
$f(0)=0$ is a local maximum and $f(1)=-1$ is a local minimum. There are no absolute extrema.
The function is concave up on $(-\infty,0)$ and on $(0,\infty)$; there is no inflection point.
See graph.
Work Step by Step
Step 1. Given the function $y=2x-3x^{2/3}$, we have $y'=2-2x^{-1/3}$ and $y''=\frac{2}{3}x^{-4/3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,1$ as critical points with $f(0)=0, f(1)=-1$
Step 3. Examine the sign change of $y'$ across the critical points: $..(+)..(0)..(-)..(1)..(+)$, we can identify $f(0)=0$ as a local maximum and $f(1)=-1$ as a local minimum. The function does not have absolute extrema and 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 $x=0$ and a possible inflection point is $(0,0)$
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 $..(+)..(0)..(+)..$,
Step 6. The function is concave up on $(-\infty,0)$ and on $(0,\infty)$, this also means that $(0,0)$ is not an inflection point (no concave change).
Step 7. See graph.
