For both, we may notice that the domain is the entire -plane and contains no boundary points. We are therefore left with points at which . For (a), this amounts to

When at one of these critical points,

and we must have . Similarly, when , we must have

and therefore . When neither nor equals , we may divide by and in both:

Multiplying both sides of the top equation by and subtracting the bottom gives

and therefore . Substitution confirms that all of these are actual solutions of . To determine the nature of each critical point, remember that the discriminant is defined as

and that

1. confirms a local extremum and

2. confirms a saddle point, while

3. tells us nothing.

Some textbooks define the discriminant as . When this happens, then confirms extrema and confirms saddle points. I'm not sure which sign your teacher uses.