Answer
Concave upward on
$$
(-\infty, 0 ) ,\quad (1, \infty ),
$$
concave downward on
$$
( 0,1 )
$$
inflection points at
$$(0, 0), \quad (1 , -3).$$
Work Step by Step
$$
f(x)=x^{\frac{8}{3}}-4 x^{\frac{5}{3}}
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{8}{3})x^{\frac{5}{3}}-4(\frac{5}{3}) x^{\frac{2}{3}}\\
&=\frac{8}{3}x^{\frac{5}{3}}-\frac{20}{3} x^{\frac{2}{3}} ,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{8}{3} (\frac{5}{3})x^{\frac{2}{3}}-\frac{20}{3}(\frac{2}{3}) x^{\frac{-1}{3}} \\
&=\frac{40}{9} x^{\frac{2}{3}}-\frac{40}{9} x^{\frac{-1}{3}}\\
\end{aligned}
$$
We factor $f^{\prime\prime}(x)$ as
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{40}{9} x^{\frac{2}{3}}-\frac{40}{9} x^{\frac{-1}{3}}\\
&=\frac{40(x-1)}{9 x^{\frac{-1}{3}}}
\end{aligned}
$$
look for an $x$-value that makes $f^{\prime\prime}(x)=0$.
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime\prime}(x) =\frac{40(x-1)}{9 x^{\frac{-1}{3}}} =0 \quad \text {when } \quad x=1
\end{aligned}
$$
but also where $f^{\prime\prime}(x)$ does not exist this happens
at $x=0$. So Check the sign of $f^{\prime\prime}(x)$ in the three intervals
$$
(-\infty, 0 ), \quad (0,1), \quad (1,\infty).
$$
(1)
Test a number in the interval $(-\infty, 0 )$ say $ -2$
$$
\begin{aligned}
f^{\prime\prime}(-2) &=\frac{40((-2)-1)}{9 (-2)^{\frac{-1}{3}}} \\
&=\frac{40\cdot \:2^{\frac{1}{3}}}{3} \\
&\gt0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty, 0 )$.
(2)
Test a number in the interval $( 0,1) $ say $ \frac{1}{2}$
$$
\begin{aligned}
f^{\prime\prime}(\frac{1}{2}) &=\frac{40((\frac{1}{2})-1)}{9 (\frac{1}{2})^{\frac{-1}{3}}} \\
&=-\frac{10\cdot \:2^{\frac{2}{3}}}{9} \\
&\lt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is negative in that interval, so $f(x)$ is concave downward on $( 0,1)$
(3)
Test a number in the interval $ (1, \infty )$ say $ 2$
$$
\begin{aligned}
f^{\prime\prime}(2) &=\frac{40((2)-1)}{9 (2)^{\frac{-1}{3}}} \\
&=\frac{40\cdot \:2^{\frac{1}{3}}}{9} \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $ (1, \infty )$.
Therefore,
Concave upward on
$$
(-\infty, 0 ) ,\quad (1, \infty ),
$$
concave downward on
$$
( 0,1 )
$$
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $ x=0 $, the graph changes from concave upward to concave downward at that point, so $(0 , f(0)$ is inflection point.
Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive
at $ x=1 $, the graph changes from concave downward to concave upward at that point, so $(1 , f(1)$ is inflection point.
Now ,we find $f( 0), f(1)$as follows:
$$
f(0)=(0)^{\frac{8}{3}}-4 (0)^{\frac{5}{3}}=0,\\
f(1)=(1)^{\frac{8}{3}}-4 (1)^{\frac{5}{3}}=-3.
$$
Finally, we have inflection points at $(0, 0), (1 , -3).$
