Answer
Local minimum:$\quad (-1,0)$ and $(1,0)$
Local maximum:$\quad$ $(0,0.75)$
Inflection points:$\quad (\pm\sqrt{3},1.19)$
Intervals where f is concave down:$\quad (-\sqrt{3},\sqrt{3})$
Intervals where f is concave up:$\quad (-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$
Work Step by Step
$y=f(x)=\displaystyle \frac{3}{4}(x^{2}-1)^{2/3}$, defined everywhere.
$f'(x)=\displaystyle \frac{3}{4}(\frac{2}{3})(x^{2}-1)^{-1/3}(2x)=\frac{x}{(x^{2}-1)^{1/3}}$
undefined for $ x=\pm 1\qquad$ ... critical points
$f'(x)=0$ for $ x=0\quad$ ... critical point
$f'(x)=x(x^{2}-1)^{-1/3}$
$f''(x)=(x^{2}-1)^{-1/3}+x[-\displaystyle \frac{1}{3}(x^{2}-1)^{-4/3}(2x)]$
$f''(x)=\displaystyle \frac{2}{3(x^{2}-1)^{1/3}}[\frac{3}{2}-\frac{x^{2}}{(x^{2}-1)}]$
$f''(x)=\displaystyle \frac{2}{3(x^{2}-1)^{1/3}}[\frac{3(x^{2}-1)-2x^{2}}{2(x^{2}-1)}]$
$f''(x)=\displaystyle \frac{x^{2}-3}{3(x^{2}-1)^{4/3}}$
$f''(x)=0$ for $x=\pm\sqrt{3}\Rightarrow(\pm\sqrt{3}, f(\pm\sqrt{3}))$ are points of inflection
$f(\pm\sqrt{3})\approx 1.19$
Using $\cap$ for "concave down" and $\cup$ for "concave up"
$\left[\begin{array}{ccccc}
interval & (-\infty,-\sqrt{3}) & (-\sqrt{3},\sqrt{3}) & (\sqrt{3},\infty)\\
t & -2 & 0 & 2\\
f''(t) & 0.077\gt 0 & -1\lt 0 & 1.56\gt 0\\
concavity & \cup & \cap & \cup
\end{array}\right]$
Apply the second derivative test (Th. 5)$\quad c=-1,0,1$
$\left\{\begin{array}{ccc}
-1\in(-\infty,-\sqrt{3}) & \Rightarrow f''(-1)\gt 0\Rightarrow\min,, & f(-1)=0\\
0\in(-\sqrt{3},\sqrt{3}) & \Rightarrow f''(0)\lt 0\Rightarrow\max, & f(0)=0.75\\
1\in(\sqrt{3},\infty) & f''(1)\gt 0\Rightarrow\min, & f(1)=0
\end{array}\right.$
