The 2nd equation gives rise to two different cases x=0 and y=0.
Case I: if x=0 then
Case II: If y=0 then,
Now we can make the Hessian matrix
Now we can check the critical points found above
So both of these are Saddle points.
Since both eigen values of the Hessian are positive the point in a local minimum.
For Part b) all we need to do is calculate the maximum and minimum on the boundary.
Since the boundary is a circle with radius , it can be parameterized by
If we put these into the original function we get
This is a function of one variable and its extreem values can be found using methods from first semester calculus.
Work not shown, but you will find that is a critical point. If you put this back into the parameteric equations you get