Answer
$$
f(x)=x^{\frac{8}{3}}+x^{\frac{5}{3}}
$$
critical numbers are: $\frac{-5}{8}$ and $0$.
By Part 2 of the second derivative test, $\frac{-5}{8}$ leads to a relative minimum.
by the first derivative test, Neither a relative minimum or maximum occurs at $0$.
Work Step by Step
$$
f(x)=x^{\frac{8}{3}}+x^{\frac{5}{3}}
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{8}{3})x^{\frac{5}{3}}+ (\frac{5}{3})x^{\frac{2}{3}}\\
&=\frac{x^{\frac{5}{3}}}{3}(8x+5)
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{x^{\frac{5}{3}}}{3}(8x+5)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
8x+5 =0 \quad &\text {or} \quad x =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
8x =-5 \quad &\text {or} \quad x =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x =\frac{-5}{8} \quad &\text {or} \quad x =0 \\
\end{aligned}
$$
critical numbers are: $\frac{-5}{8}$ and $0$.
Now use the second derivative test. The second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x)&=\frac{8}{3} (\frac{5}{3})x^{\frac{2}{3}}+ \frac{5}{3} (\frac{2}{3})x^{\frac{-1}{3}} \\
&=\frac{40}{9} x^{\frac{2}{3}}+ \frac{10}{9}x^{\frac{-1}{3}}.
\end{aligned}
$$
Evaluate $f^{\prime\prime}(x)$ first at $x=\frac{-5}{8}$, getting
$$
\begin{aligned}
f^{\prime\prime}(\frac{-5}{8})&=\frac{40}{9} (\frac{-5}{8})^{\frac{2}{3}}+ \frac{10}{9}(\frac{-5}{8})^{\frac{-1}{3}}\\
&=\frac{2\cdot \:5^{\frac{2}{3}}}{3}\\
& \approx 1.94934 \gt 0
\end{aligned}
$$
so that by Part 1 of the second derivative test, $\frac{-5}{8}$ leads to a relative minimum.
Now, evaluate $f^{\prime\prime}(x)$ at $x=0$, getting
$$
\begin{aligned}
f^{\prime\prime}(0)&=\frac{40}{9} (0)^{\frac{2}{3}}+ \frac{10}{9}(0)^{\frac{-1}{3}}\\
& f^{\prime\prime} \text{ does not exist}
\end{aligned}
$$
so the test gives no information about extrema, we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(\frac{-5}{8}, 0 ), \quad (0,\infty).
$$
(1)
Test a number in the interval $(\frac{-5}{8}, 0 )$ say $\frac{-1}{2}$:
$$
\begin{aligned}
f^{\prime}(\frac{-1}{2}) &=(\frac{8}{3})(\frac{-1}{2})^{\frac{5}{3}}+ (\frac{5}{3})(\frac{-1}{2})^{\frac{2}{3}} \\
&=\frac{1}{3\cdot \:2^{\frac{2}{3}}}\\
&\approx 0.20998
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(\frac{-5}{8}, 0 )$
(2)
Test a number in the interval $(0, \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=(\frac{8}{3})(1)^{\frac{5}{3}}+ (\frac{5}{3})(1)^{\frac{2}{3}} \\
&=\frac{13}{3}\\
&\approx 4.33333
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is
also increasing on $(0, \infty ).$
So that by the first derivative test, Neither a relative minimum or maximum occurs at $0$.
