= Solution
For $x>0$, differentiate the two explicit branches:
$$
y_\pm'=x\left(2\pm\frac52\sqrt x\right),\qquad
y_\pm''=2\pm\frac{15}{4}\sqrt x.
$$
The plus branch is strictly increasing and strictly <convex> for $x>0$. Its only zero is $x=0$, also its one-sided <global minimum>; it has no interior extremum or <inflection point>.
The minus branch rises initially, then has $y_-'=0$ at $\sqrt x=4/5$, changing from positive to negative. Its second <derivative> changes sign at $\sqrt x=8/15$. Its zeros follow from $y_-=x^2(1-\sqrt x)$. Consequently
$$
\boxed{\begin{array}{c|c|c}
&y_+&y_-\\\hline
\text{zeros}&0&0,\ 1\\
\text{interior maximum}&\text{none}&16/25\\
\text{interior minimum}&\text{none}&\text{none}\\
\text{inflection}&\text{none}&64/225
\end{array}}
$$
The origin is a one-sided <local minimum> of both graph branches, because their domain ends there; it is not an interior <critical point>. The minus branch has no <global minimum> since it tends to $-\infty$ as $x\to\infty$. Its maximum and inflection heights, useful for the sketch, are $256/3125$ and $28672/759375$, respectively.
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