Answer
Critical points $x= 0, 12/5, 3$, endpoint $x=3$.
At $x=0$, $y=0$ minimum.
At $x=12/5$, $y=\frac{144}{125}\sqrt {15}$ local maximum.
At $x=3$, $y=0$ minimum.
Work Step by Step
Step 1. Given the function $y=x^2\sqrt {3-x}=x^2(3-x)^{1/2}$, we can find its derivative as $y'=2x(3-x)^{1/2}+\frac{1}{2}x^2(3-x)^{-1/2}(-1)=\frac{2x(3-x)}{\sqrt {3-x}}-\frac{x^2}{2\sqrt {3-x}}=\frac{12x-5x^2}{2\sqrt {3-x}}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x= 0, 12/5, 3$. The domain of the function is $(-\infty,3)$ and the endpoint is $x=3$.
Step 3. At $x=0$, $y=0$ and we can identify this as a minimum as shown in the figure.
Step 4. At $x=12/5$, $y=(12/5)^2\sqrt {3-12/5}=\frac{144}{25}\sqrt {\frac{3}{5}}=\frac{144}{125}\sqrt {15}$ and we can identify this as a local maximum (use local test points as necessary) as shown in the figure.
Step 5. At $x=3$, $y=0$ and we can identify this as a minimum as shown in the figure.