Existence of Partial Derivatives Implies Differentiability Doing some self-study. My textbook has this Theorem (see below). I understand it but was hoping for something shorter and more intuitive. Any thoughts?

 A: There are no shortcuts here, you need to follow the proof and understand it. Indeed differentiability and continuity issues for functions of several variables are less intuitive than for function of one variable. For example, the existence of all directional derivatives at a point does not imply continuity whereas for function of one variable derivability implies continuity.
Therefore maybe it could be useful make a general comment here on the topic of Differentiability and Continuity for functions of several variables.
1. If a function is discontinuos then it can't be differentiable, infact continuity is a necessary condition since differentiability implies continuity.
2. If partial derivatives do not exist then $f$ can't be
        differentiable, infact existence of partial derivatives is a necessary condition since differentiability implies their existence.
3. If partial derivatives exist and are continuos you are done, infact for the "Differentiability theorem" if all the partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.
4. If partial derivatives are not continuos at the point you can't yet
    conclude anything about differentiability. You need to check
    directly differentiability by definition that:
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$
A: It is hard to imagine a shorter or more intuitive proof. The proof consists of writing the derivative, and then checking directly that it satisfies the definition of derivative, using direct estimates that follow from the hypotheses. 
