Answer
At $x=0$, $y=1$, local minimum
Work Step by Step
Step 1. Given the function $y=\frac{1}{\sqrt[3] {1-x^2}}=(1-x^2)^{-1/3}$, we can find its derivative as $y′=-\frac{1}{3}(1-x^2)^{-4/3}(-2x)=\frac{2x}{3\sqrt[3] {(1-x^2)^4}}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=0$ or $1-x^2-0$ and we have $x=0, \pm1$
Step 3. At $x=0$, $y=1$ and we can identify this as a local minimum (use local test points) as shown in the figure.
Step 4. At $x=\pm1$, $y=\pm\infty$ and we can identify that there are asymptotes at these locations, not extrema.
