min/max/saddle points of $z=\frac{1}{2} -\sin^2(x^2+y^2)$ 
$$z=\frac{1}{2} -\sin^2(x^2+y^2)$$ in the interval $0\leq x^2+y^2 \leq \frac{\pi}{6}$

$z_{x}:-2x\sin(2x^2+2y^2)=0$ so $x=0$ or $(2x^2+2y^2)=0$
$z_{y}: -2y\sin(2x^2+2y^2)=0$ so $y=0$ or $(2x^2+2y^2)=0$
$z_{xx}: -2\sin(2x^2+2y^2)-8x^2\cos(2x^2+2y^2)<0$ due to the domain
$z_{yy}: -2\sin(2x^2+2y^2)-8y^2\cos(2x^2+2y^2)<0$ due to the domain
D=$z_{xx}\cdot z_{yy}-[z_{xy}]^2=4\sin(2x^2+2y^2)+16y^2\cos(2x^2+2y^2)\sin(2x^2+2y^2)+16x^2\cos(2x^2+2y^2)\sin(2x^2+2y^2)$
I do not see why $(2x^2+2y^2)=0$ can not be a point and why $D>0$
And why it is maximum
the answers: 1 and 2

I really do not know why Wolfram Alpha gives you result which states that there are four local maxima because it seems to me that I can elementarily prove that there is only one local maximum, if someone finds a flaw in my stream of thought let me know about it.
First, it is clear that $(0,0)$ is a local maximum.
Suppose now that there exists some local maximum $(x_0,y_0) \neq (0,0)$. That would mean that there exists some $\epsilon_0$ such that we have $z(x_0,y_0) \geq z(x,y)$ for every $(x,y)$ inside the ball of radius $\epsilon_0$ that has the center at the point $(x_0,y_0)$.
So it means that we would have $\frac{1}{2} -\sin^2(x_0^2+y_0^2) \geq \frac{1}{2} -\sin^2(x^2+y^2)$ for every $(x,y)$ in the ball with center at the point $(x_0,y_0)$ with radius $\epsilon_0$.
So we would have $\sin^2(x_0^2+y_0^2) \leq \sin^2(x^2+y^2)$ for all $(x,y)$ inside that ball.
Now choose some point $(x_1,y_1)$ that lies in the ball which is such that we have $x_1^2 + y_1^2 < x_0^2 + y_0^2$ (it is clearly obvious that such a point exists). This implies that $\sin(x_1^2+y_1^2) < \sin(x_0^2+y_0^2)$ which implies $\sin^2(x_1^2+y_1^2) < \sin^2(x_0^2+y_0^2)$, so we have arrived at the contradiction and the only maximum is $(0,0)$.
If it is not clear to you that $(0,0)$ is a local maximum observe that for every pair $(x,y)$ which satisfies $0 < x^2+y^2 \leq \frac{\pi}{6}$ We have $0 < \sin^2(x^2+y^2) \leq \frac {1}{4}$ and for $(0,0)$ we have $\sin^2(0^2+0^2)=0$.
