Can "being differentiable" imply "having continuous partial derivatives"? Consider the following theorem:

Let $E$ be a subset of ${\bf R}^n$, $f:E\to {\bf R}^m$ be a function, $F$ be a subset of $E$, and $x_0$ be an interior point of $F$. If all the partial derivatives $\frac{\partial f}{\partial x_j}$ exist on $F$ and are continuous at $x_0$, then $f$ is differentiable at $x_0$.

And I consider the converse of the above theorem:  

Let $E$ be a subset of ${\bf R}^n$, $f:E\to {\bf R}^m$ be a function, $F$ be a subset of $E$, and $x_0$ be an interior point of $F$. If $f$ is differentiable at $x_0$, then all the partial derivatives $\frac{\partial f}{\partial x_j}$ exist on some neighbourhood of $x_0$ and are continuous at $x_0$.

It is trivial to show that the converse is NOT true when $m=1$. It seems no hope that it will be true when $m\geq 2$. Here is my question:

Is the converse
  true when $m\geq 2$? If is not true, how to construct the counterexample?

Edit: The title is corrected. 
Edit: Since another question is not relevant to the first one here, I think, I put it into another post.
 A: Re your first question, the converse is false and one most usually mentions counterexamples based on the oscillatory behaviour of $x\mapsto\sin(1/x)$ near $x=0$.
Re your second question, Chern's and Folland's definitions are identical since both require $D^kf$ to be continuous for $f$ to belong to $C^k$, as they should.
Edit Due to a radical modification of the question, the above only partly applies to the current version of the question. Great...
However, one can say this: the case $m\ge2$ cannot yield stronger results, simply consider a function like $(f,0,0,\ldots,0)$ where $f$ is a counterexample when $m=1$.
A: No, there are differentiable functions which are not $C^1.$ For example consider $f:\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2 \sin(\frac{1}{x})$ for $x \neq 0$ and $f(x) = 0$ if $x = 0.$ Then $f$ is differentiable but not $C^1.$
To see that $f$ is differentiable away from zero is not hard and follows from the product and chain rules. On the other hand to show $f$ is diffentiable at $0$ we must appeal to the definition of the derivative. 
Observe
$$\displaystyle\lim_{x\rightarrow 0} \frac{f(x) - f(0)}{x-0} = \displaystyle\lim_{x\rightarrow 0} \text{  } x \sin(1/x)$$ 
which converges to $0$ as $|x \sin(1/x)| < |x|$.
Hence, $f$ is everywhere diffentiable.
It follows $f'(x)$ is given by $2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})$ if $x \neq 0$ and $0$ if $x = 0.$
We claim $f'$ is not continuous at $0.$ To see this, note that if $f'$ were continuous at $0,$ then so too would be the function $g(x)$ defined by $\cos(1/x)$ if $x \neq 0$ and $0$ if $x = 0.$ But this is false as $\langle\frac{1}{\pi + n2\pi}\rangle_{n\in\mathbb{Z}_+}$ is a sequence of real numbers converging to $0$ such that 
$$\displaystyle\lim_{n\rightarrow \infty} g\left(\frac{1}{\pi + n 2\pi}\right) = 1 \neq 0 = g(0).$$   
Hence, $f$ is differentiable but not $C^1.$
A: Here is a counterexample from an online article:
$$
f(x,y)=\begin{cases}
(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}},&(x,y)\neq(0,0)\\
0,&(x,y)=(0,0).
\end{cases}
$$
A: The following theorem in baby Rudin can answer the question:

