differentiability and continuity of multivariable functions I'm trying to understand differentiability of multivariable functions.
The textbook says,
"If the partial derivatives ƒx and ƒy of a function ƒ(x, y) are continuous throughout an open region R, then ƒ is differentiable at every point of R."
Hass, Joel R.; Heil, Christopher E.; Weir, Maurice D.. Thomas' Calculus (Page 818). Pearson Education. Kindle Edition.
So in two dimensions, if something is continuous, it might not be differentiable, because it could be pointy
(that's an official math term, right?) Couldn't that happen in three dimensions too?
Also, I was wondering whether the converse of the above is true - i.e. if a multivariable function is differentiable, that means it's continuous and that the partial derivatives exist. And if not, what's the counterexample?
Thank you!
 A: No, “pointy” is not an official math term. Not only there is no such thing as official math terms, as I have never seen it on a mathematical text.
But, yes, just like in the case of functions from $\mathbb R$ into $\mathbb R$, a continuous functions may fail to be differentiable. An example would be$$\begin{array}{ccc}\mathbb R&\longrightarrow&\mathbb R\\x&\mapsto&\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$By the way, the graph of this function is not “pointy”.
And, yes, this can also happens in the context of function from $\mathbb R^n$ into $\mathbb R$.
On the other hand, asserting that a function $f$ is differentiable does not mean that $f$ is continuous and that the partial derivatives exists. It is stronger than that. An example would be$$\begin{array}{rccc}f\colon&\mathbb R^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&\begin{cases}\frac{xy}{x^2+y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$It has partial derivatives everywhere, but it is not differentiable at $(0,0)$.
A: Note that the book does not state that $f$ needs to be continuous, but rather that the partial derivatives are continuous. So you actually misinterpreted what the converse of the  theorem would be. 
The converse would be that, if $f$ is differentiable, then the partial derivatives exists and are continuous.
The first part of the statement is true, the partial derivatives do exist if $f$ is differentiable, but the partial derivatives are not necessarily continuous, not even in the one dimensional case. The classic counter example would be
$$x \mapsto \begin{cases} x^2 sin(\frac{1}{x}) &, x\neq 0 \\ 0 & ,x=0 \end{cases}$$
For more examples see Discontinuous derivative.
