Find the critical points of $f (x, y) = 2xy$ when restricted to $$ x^2 + 2y^2 = 6$$
I just wanna make sure I understand one thing, in the answer they provided the values they got after doing lagrange. But just to confirm so I give right answers on an exam, the only reason it is considered to be critical point is because of "=" sign in $ x^2 + 2y^2 = 6$ right? If it had been "$\leq$" as in $ x^2 + 2y^2 \leq 6$ then it would have been considered an edge point and not a critical right? The reason I'm asking this is because in another similar question, where it was an inequality, they only gave the points when gradient of $f$ was zero when asked for critical points.
Btw I don't help solving this problem.
You're basically correct.
Remember that a critical point of a function is one where the derivative of the function is either equal to zero, or it is undefined. And that is going to depend on the domain of the function.
If the domain of $f$ is the curve $x^2 + 2y^2 = 6$, then its derivative is defined only in one direction, which is tangent to that ellipse (in fancy terms, its domain is a one-dimensional manifold). If the domain instead is the region $x^2 + 2y^2 \leq 6$, then its derivative can be considered in multiple directions that can be separated into two components (because the domain is now a two-dimensional manifold). As such, having a zero derivative in one direction doesn't force you to have a zero derivative in another direction (and, in fact, for smooth functions you'll always have a direction where the derivative is at a maximum and another direction where the derivative is zero).
So yes, a critical point of "$f$ on the boundary" is not necessarily a critical point of "$f$ on the whole region". It might be, but it's not guaranteed.