Answer
no local extrema, no absolute extrema;
concave up on $(-\infty,-1), (0,\infty)$ and concave down on $(-1,0)$;
$x=-1,0$ are inflection points.
See graph.
Work Step by Step
Step 1. Given the function $y=\sqrt[3] {x^3+1}=(x^3+1)^{1/3}$, we have $y'=\frac{1}{3}(x^3+1)^{-2/3}(3x^2)=x^2(x^3+1)^{-2/3}$ and $y''=2x(x^3+1)^{-2/3}+\frac{-2x^2}{3}(x^3+1)^{-5/3}(3x^2)=\frac{2x(x^3+1)}{\sqrt[3] {(x^3+1)^5}}-\frac{2x^4}{\sqrt[3] {(x^3+1)^5}}=\frac{2x}{\sqrt[3] {(x^3+1)^5}}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=-1,0$ as critical points with $y(-1)=0, y(0)=1$.
Step 3. Examine the sign change of $y'$ across the critical points: $..(-1)..(+)..(0)..(+)..$ As $y'\geq0$, we do not have any local extrema. As $x\to\pm\infty, y\to\pm\infty$, there are no absolute extrema.
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 within the domain.
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 up on $(-\infty,-1), (0,\infty)$ and concave down on $(-1,0)$, this means that $x=-1,0$ are inflection points.
Step 7. See graph.
