Answer
Local minimum:$\quad (1,-\displaystyle \frac{27}{7})$
Local maximum:$\quad$ $(-1,\displaystyle \frac{27}{7})$
Inflection points:$\quad (0,0)$
Intervals where f is concave down:$\quad (-\infty,0)$
Intervals where f is concave up:$\quad (0,\infty)$
Work Step by Step
$y=f(x)=\displaystyle \frac{9}{14}x^{1/3}(x^{2}-7)=\frac{9}{14}x^{7/3}-\frac{9}{2}x^{1/3}$,
which is defined everywhere.
$f'(x)=\displaystyle \frac{9}{14}(\frac{7}{3})x^{4/3}-\frac{9}{2}(\frac{1}{3})x^{-2/3}$
$f'(x)=\displaystyle \frac{3x^{4/3}}{2}- \frac{3}{2x^{2/3}} = \frac{3x^{2}-3}{2x^{2/3}} = \frac{3(x^{2}-1)}{2x^{2/3}}$
which is undefined for $ x=0\qquad$ ... critical point
$f'(x)=0$ for $ x=\pm 1\quad$ ... critical points.
$f'(x)=\displaystyle \frac{3}{2}x^{2/3}-\frac{3}{2}x^{-2/3}$
$f''(x)=\displaystyle \frac{3}{2}\cdot\frac{2}{3}x^{-1/3}-\frac{3}{2}(-\frac{2}{3})x^{-5/3}$
$f''(x)= \displaystyle \frac{1}{x^{1/3}} + \frac{1}{x^{5/3}} = \frac{x^{2}+1}{x^{5/3}}$
$f''$ has no zeros and is undefined at $x=0$, the only critical point.
On $(-\infty,0),\ f''$ is negative, f is concave down,
and there is a local maximum at $x=-1 \in(-\infty,0)$.
$f(-1)=\displaystyle \frac{27}{7}.$
On $(0,\infty),\ f''$ is positive, f is concave up,
and there is a local minimum at $x=1 \in(-\infty,0)$.
$f(1)=-\displaystyle \frac{27}{7}.$
At $(0,0)$, the concavity changes: inflection point.
