First derivative, critical numbers, and classification of extrema

Rewrite $h(x) = x^{5/3} - 5x^{2/3}$. Differentiating term by term gives $h'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}(x - 2)$. This is undefined at $x = 0$ (since it involves $x^{-1/3}$) and equals zero when $x = 2$. Since $h(0) = 0$ is defined, both $x=0$ and $x=2$ are critical numbers.

To build a sign chart, test values in each interval determined by $x=0$ and $x=2$: for $x<0$, $x^{-1/3}<0$ and $(x-2)<0$, so $h'(x)>0$; for $0<x<2$, $x^{-1/3}>0$ and $(x-2)<0$, so $h'(x)<0$; for $x>2$, both factors are positive, so $h'(x)>0$. Thus $h$ is increasing on $(-\infty,0)$, decreasing on $(0,2)$, and increasing on $(2,\infty)$.

Since $h'$ changes from positive to negative at $x=0$, $h$ has a local maximum there, with $h(0)=0$. Since $h'$ changes from negative to positive at $x=2$, $h$ has a local minimum there, with $h(2) = 2^{2/3}(2-5) = -3\cdot 2^{2/3}$.

That critical number is x equals zero, where the derivative is undefined due to the negative exponent term. The sign chart shows the derivative changes from positive to negative there, since h is increasing just before zero and decreasing just after. By the First Derivative Test, this sign change from positive to negative means x equals zero is a local maximum.