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
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.