$\nabla f=0$ implies $f$ constant 

For the maroon underline, I thought that the existence of partial derivatives did not imply f was differentiable?
For the green part I can't fathom which theorem the lecturer is talking about, so I don't follow this part either.

Ummm. If you have a point $(a,b)$ in the plane, say, and another $(c,d),$ the fact that the $x$ derivative stays $0$ means that $f(a,b) = f(c,b).$ Then, the fact that the $y$ derivative remains $0$ means that $f(c,b) = f(c,d).$ Together, $f(a,b) = f(c,d).$ The argument is the same if the number of variables is larger.
I see, there is a comment to this effect by Nameless, having been deleted as an answer. Well, there you go. 
