Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich

Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$

What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{\partial z}{\partial y} = x \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
And I've tried to simplify, and I got:
$$\frac{y a^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{x b^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
Now the problem is, some cubic terms appear and I don't know how to solve this and find the critical points to proceed with the exercise. Of course the denominator cannot be zero, so my system is composed only by the numerators.
I'm pretty sure I'm doing something wrong, 'cause this seems too much trouble for an introductory exercise (it's one of the firsts in the chapter)
 A: I suppose that $a,b>0$ are given parameters. The given function is then defined in the region 
$$
R = \left\{\ (x,y)\ :\ \frac {x^2}{a^2}+ \frac {y^2}{b^2}\le 1\ \right\}\ .
$$
The four quadrants $I, II, \dots$ are cutting $R$ in four pieces, $R_I$, $R_{II}$, ... and it is enough to search for the local / global maxima on the piece $R_I$. 
On the boundary of $R_I$ (and of $R$) the given function vanishes.
The (local/global) maximal values of the function $f$ on $R_I$ correspond to the (local/global) maximal values of $f^2/(a^2b^2)$, which is a function of $X=x^2/a^2$, $Y=y^2/b^2$, explicitly given by the help function
$$
h(X,Y)=XY\left(1-X-Y\right)=XY-X^2Y-XY^2\ ,
$$
the corresponding region for $(X,Y)$ is a triangle given by $X,Y,1-X-Y\ge 0$. At the boundary the values are zero. Differentiation gives the system of equations for a local extremal point:
$$
\left\{
\begin{aligned}
Y&= 2XY +Y^2\ ,\\
X&= 2XY +X^2\ ,
\end{aligned}
\right.
$$
and subtraction gives $Y-X=Y^2-X^2=(Y-X)(Y+X)$. We have thus either $X=Y$, or $X+Y=1$.


*

*in case of $X=Y$ the common value is $0$ or $1/3$.

*in case of $X+Y=1$ adding the equations we get $1=X+Y=4XY+X^2+Y^2=2XY+1^2$, so $XY=0$, a boundary point.


So we keep only $(X,Y)=(1/3,\;1/3)$.
A: With the polar coordinates $\frac xa = r\cos t$ and $\frac yb  =r \sin t$,
$$z=\frac12ab r^2\sin 2t \sqrt{1-r^2}$$
For any given $r\le 1$, 
$$|z(r)|\le \frac12abr^2 \sqrt{1-r^2}$$
Then, set $(r^2 \sqrt{1-r^2})'=0$ to locate the extreme at $r^2=\frac23$. Thus, 
$$-\frac{ab}{3\sqrt3}\le z \le \frac{ab}{3\sqrt3}$$
