Answer
See graph and explanations.
Work Step by Step
Step 1. See the graph for the function $y=x^{2/3}-(x-1)^{1/3}$.
Step 2. We have $y'=\frac{2}{3}x^{-1/3}-\frac{1}{3}(x-1)^{-2/3}$
Step 3. Letting $y'=0$, we can solve the equation $8x^2-17x+8=0$ to get $x_{\pm}=\frac{17\pm\sqrt {33}}{16}$ with $x_-\approx0.7, x_+\approx1.4$. Thus the critical points are $x=0,1,x_\pm$ including points where $y'$ are undefined.
Step 4. Checking the signs of $y'$ across the critical points, we have $..(-)..(0)..(+)..(x_-)..(-)..(1)..(-)..(x_+)..(+)..$; thus the function decreases over $(-\infty,0),(x_-,1),(1,x_+)$ and increases over $(0,x_-), (x_+,\infty)$. And $x=x_-$ gives a local maximum and $x=0,x_+$ gives local minima.
Step 5. As $x\to0^-, y'\to-\infty$, $x\to0^+, y'\to\infty$, the function has a cusp at $x=0$ where $y'(0)$ is undefined.
Step 6. As $x\to1^-, y'\to\infty$, $x\to1^+, y'\to\infty$, the function has a vertical tangent line at $x=1$ where $y'(1)$ is undefined.
