Answer
See explanations.
Work Step by Step
Step 1. Given the first derivative $y'=x^{-2/3}(x-1)=x^{1/3}-x^{-2/3}$, we have $y''=\frac{1}{3}x^{-2/3}+\frac{2}{3}x^{-5/3}=\frac{1}{3}x^{-5/3}(x+2)$
Step 2. Possible critical points are at $x=0,1$. Determine the increasing ($y'\gt0$) and decreasing ($y'\lt0$) regions by signs of $y'$: $..(-)..(0)..(-)..(1)..(+)..$; thus the function decreases on $(-\infty,0),(0,1)$ and increases on $(1,\infty)$. A local and absolute minimum can be found at $x=1$. There is no local or absolute maximum.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=-2,0$: $..(+)..(-2)..(-)..(0)..(+)..$. Thus, the function is concave up on $(-\infty,-2)$,$(0,\infty)$ and concave down on $(-2,0)$. Thus $x=-2$ and $x=0$ are inflection points.
Step 4. Sketch the function based on the above results as shown.
