$f(x,y)=\dfrac{\sqrt{9-x^2-y^2}}{\log(x^2+y^2-7)}$
i.Determine the maximum domain and draw it.
ii.Determine the subset of $D$ where $f$ is differentiable and compute the gradient.
iii.Find the possible points of local max and local min of the function $f$ in the given region $(x^2-y^2>8)$
For solnving first two steps i used polar system and get
$f(x,y)=\dfrac{\sqrt{9-r^2}}{\log(r^2-7)}$
But I absolutely stopped on the third part any help are welcome.
With the second point solved, the third is immediate. Anyway, the function doesn't depend on the polar angle, so, we can expect that the possible stationary points form circles around the origin.
$$\dfrac{\partial f}{\partial r}=-\dfrac{2r(9-r^2)+r(r^2-7)\log(r^2-7)}{(r^2-7)\sqrt{9-r^2}\log^2(r^2-7)}=0$$
The condition $x^2-y^2\gt8$ implies that $x^2\gt8$ and $r^2\gt8$, so we can find the values for $r$ and later apply the restriction given if necessary.
We have to find $r$ for wich the numerator vanishes. The denominator is not a problem as $8\lt r^2\lt9$ and cannot be zero. So is, consider the function,
$$d(r)=2r(9-r^2)+r(r^2-7)\log(r^2-7)$$
But it doesn't vanish as for the range considered $9-r^2\gt0$ and $r^2-7\gt0$
So, There are no stationary points.