How to properly find a critical point for $xy^2$? Let $f(x,y) = xy^2$ and the domain $D = \lbrace (x,y)| x,y\geq0, x^2 + y^2 \leq 3 \rbrace$
$f_x(x,y) = y^2$ and $f_y(x,y) = 2xy$ 
Therefore, the critical points should be $\lbrace (x,y)| y = 0, \sqrt{3} \geq x \geq 0 \rbrace$. 
The determinant of the Hessian is 
$$\det(Hf(x,y))=
\begin{vmatrix}
0 & 2y \\ 
2y & 2x
\end{vmatrix} = 0-4y^2$$
But this doesn't make sense to me because, if $y = 0$, then $f(x,0) = 0$. But on this interval, suppose I chose $(x,y) = (1,\sqrt{2})$, this would be larger and would be the maximum value of the system.  
Why didn't finding the first partials and the Hessian allow me to compute the critical point? 

On the circle of radius $r$, we have $x^2+y^2=r^2$. Therefore,
$$
xy^2=r^2x-x^3\tag1
$$
This implies that the interior critical points are at
$$
(x,y)=\frac r{\sqrt3}\left(\pm1,\pm\sqrt2\right)\tag2
$$
with the corresponding values of
$$
xy^2=\pm r^3\frac2{3\sqrt3}\tag3
$$
At the endponts of $x=\pm r$, we get the values of $xy^2=0$. Therefore,
$$
-r^3\frac2{3\sqrt3}\le xy^2\le r^3\frac2{3\sqrt3}\tag4
$$
Thus, the maxima and minima are on the bounding circle.
If $x^2+y^2=3$, then $r=\sqrt3$ and so $(4)$ says that
$$
-2\le xy^2\le2\tag5
$$
