Are there any functions that are differentiable but not continuously-differentiable? Let $U$ be an open set on ${\mathbb R}^{n}$ (but $U$ is not an empty set), $\textbf{p}\in{U}$, and $f:U\to \mathbb R$ is continuously-differentiable on $U$. Then, it is known that, "the function $f$ can be differentiable for all $\textbf{q}\in U$." (See Spivac)
And I know that, there is a function $f$ such that it is differentiable at $\textbf{p}$ but, for any $r> 0$,  $f$ is not differentiable  (and continuously differentiable) on 
$U_{\textbf{p}} (r)$ .
Here, $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on 
$\textbf{p}$.
For example, if $f:{\mathbb R}^{2}\to \mathbb R$ is defined as follows, $f$ is differentiable at $\textbf{0}$,, but is not differentiable (and not continuous) at any other point. Here $\mathbb Q$ is the set of all rational numbers, and $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on $\textbf{p}$. 
$f(x,y):=\left\{
\begin{array}{rr}
0, &  (x,y)\in \mathbb Q^{2} \\
x^2 + y^2, &  (x,y)\notin \mathbb Q^{2} \\
\end{array}
\right.$

Therefore, there is at least one function that does not have a continuously-differentiable region, even if it can be differentiable at one point. 
But I cannot imagine whether are there any functions that are differentiable on $U$ but not continuously-differentiable.

My question
  Let $U$ be an open set of $\mathbb R^n$ (but is not an empty set), and $\  \textbf{p}\in U $.
  Then, are there any functions  $f:U\to \mathbb R$  such that,
  $f$ is differentiable on $U$, but for any $r> 0$,  $f$ is not continuously-differentiable on $U_{\textbf{p}} (r)$ ? 
  If so, give an example. If not, please explain why.
  Here, $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on $\textbf{p}$.

Here, the definitions of differentiable and continuously differentiable are as follows.

Def1 (Differentiable at $\textbf{p}$) 
  Let $U$ be an open set (but not empty set) of ${\mathbb R}^{n}$, $\textbf {p} \in \mathbb R^n$, and $f$ is a function whose domain is $U$.
  At this time, $f$ is differentiable at $\textbf{p}$ iff the following is satisfied.
${\exists} A:{\mathbb R}^{n}\to \mathbb R$: a linear map such that 　
   $${\lim}_{\textbf{x}\to\textbf{p}}\frac{|f(\textbf{x}) - A(\textbf{x}-\textbf{p}) - f(\textbf{p})|}{|\textbf{x}-\textbf{p}|} = 0$$



Def2 (Differentiable on $\textbf{U}$) 
  Let $U$ be an open set (but not empty sets) of ${\mathbb R}^{n}$, and $f$ is a function which domain is $U$.
  At this time, $f$ is differentiable at $U$ iff "for all $\textbf{q}\in{\mathbb R}^{n}$, $f$ is differentiable at  $\textbf{q}$".



Def3 (Continuously-differentiable on $U$) 
  Let $U$ be an open set (but is not empty sets) of ${\mathbb R}^{n}$, and $f$ is a function which domain is $U$.
  At this time, $f:U\to \mathbb R$ is continuously-differentiable on $U$ iff 
  
  
*
  
*$f$ is  partially differentiable for all direction, ${x}_{1},
   {x}_{2}, ..., {x}_{n}$ (that mean, we can define $\frac{\partial f}{\partial{x}_{1}}, \cdots\frac{\partial f} {\partial{x}_{n}} $ on
  $U$).  and,
  
*$\frac{\partial f} {\partial{x}_{1}}, \cdots\frac{\partial f}
   {\partial{x}_{n}} $ are continuous on $U$.
  

P.S.
I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions.

Post-hoc Note:【Verification of the function taught by Thomas Shelby】
The following are the confirmation that the following function $f$ meets my requirement (Is it correct as proof?):
$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac1 {\sqrt{x^2+y^2}}\right),&(x,y)\neq 0\\0,&(x,y)=0\end{cases}.$

My proof:
${\lim}_{\|\textbf{x}\|\to 0} \frac{f(\textbf{x}) - f(\textbf{0})}{\|\textbf{x}\|}=
{\lim}_{\|\textbf{x}\|\to 0} \frac{{\|\textbf{x}\|}^{2}\sin(1/ \|\textbf{x}\| - 0)}{\|\textbf{x}\|}=
$
${\lim}_{\|\textbf{x}\|\to 0} \|\textbf{x}\|\sin(1/ \|\textbf{x}\|)
= 0$
  Therefore, the $f$ is differntiable at $(0,0)$ and $Jf(0,0)=(0,0)$.
On the other hand, for $\textbf{x}\neq\textbf{0}$, 
  Let $g$  and $h$ be $g(x,y):=\sqrt{{x}^2 + {y}^2}\ $ and $\ h(t):={t}^{2}\sin(1/t)$ (for $t\neq 0$) respectively, then
$$\frac{d\sqrt{t}}{dt} = \frac{1}{2\sqrt{t}} $$ and,
  $$(J\|\textbf{x}\|^2)(x,y) = (2x,2y) ,$$  Therefore,
  $$(Jg)(x,y) = \left(\frac{x}{\|\textbf{x}\|} , \frac{y}{\|\textbf{x}\|}\right)\quad (\textrm{for all $\textbf{x}\neq\textbf{0}\ $}),$$ and
  $$\ \frac{d\sin(1/t)}{dt} = -\frac{\cos(1/t)}{t^2}\ \ \ (\textrm{at $t\neq 0$}).$$  Therefore,
  $$\frac{dh}{dt} ={t}^{2}\frac{d\sin(1/t)}{dt} + 2t\sin(1/t) = -\cos(1/t) + 2t\sin(1/t).$$
Therefore, at $\textbf{x}\neq \textbf{0}$, 
$$Jf(x,y) = \left(\left.\frac{dh}{dt}\right|_{t=||\textbf{x}||}\right)(Jg)(x,y) = (-\cos(1/||\textbf{x}||) + 2t\sin(1/||\textbf{x}||))\left(\frac{x}{||\textbf{x}||} , \frac{y}{||\textbf{x}||}\right).$$
Therefore,
$$\frac{\partial f}{\partial x} =  -\frac{x\cos(1/||\textbf{x}||)}{||\textbf{x}||} + 2x\sin(1/||\textbf{x}||),\,\textbf{x}\neq\textbf{0}$$and
  $$\frac{\partial f}{\partial y} =  -\frac{y\cos(1/||\textbf{x}||)}{||\textbf{x}||} + 2y\sin(1/||\textbf{x}||),\,\textbf{x}\neq\textbf{0}.$$
However, both $\dfrac{x\cos(1/||\textbf{x}||)}{||\textbf{x}||} $ and 
   $\dfrac{y\cos(1/||\textbf{x}||)}{||\textbf{x}||} $ do not get converted at $(0,0)$.
So, both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not continuous at $(0,0)$. ■

 A: An example for $n = 1$ from the theory of random walks.  Let $f$ be a (-n everywhere) discontinuous Lebesgue measurable function on $\mathbb{R}$.  Here's an example with $f$ bounded by $1$, just showing the part $x \in [-3,3]$.  (Note that I have only barely subsampled the graph in this interval.  If I were to fully sample it, this finite resolution representation would almost surely appear to be a solid rectangle of points of the graph.  Actually produced by generating $10^6$ uniformly distributed reals in $[-1,1]$ assigned to evenly spaced abscissae, then plotting a subsample of size $10^4$.)

This function is almost surely nowhere continuous (as any open interval almost surely contains points of heights arbitrarily close to $-1$ and $1$).  The integral of this function, 
$$  \int_{0}^x \; f(t) \,\mathrm{d}t  $$
is differentiable, but there's no hope of continuous differentiability.  Graph of the integral (actually, Riemann sum approximations using $10^6$ intervals in $[-3,3]$):

Picking a different instance of a bounded by $1$ discontinuous Lebesgue measurable function on $\mathbb{R}$ and integrating it the same way, we can graph the integral.

These are almost everywhere differentiable by construction (by Lebesgue's differentiation theorem); we know the derivative is $f$.  (The theorem generalizes to $n > 1$ and the integral to $\int_{[0,x_1]\times [0,x_2] \times \cdots \times [0,x_n]} \; f(t) \,\mathrm{d}t$ where we understand the intervals to be $[0,a]$ when $0 \leq a$ and $[a,0]$ when $a < 0$.)  In some way, "most" functions are everywhere discontinuous messes, so "most" functions can be integrated to a differentiable, but not continuously differentiable, function.
(This construction can be iterated to get a function that is several times continuously differentiable, but whose "last" derivative is not continuous.)
A: Consider  $f:\mathbb R^2\to \mathbb R $ defined by
$$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac1 {\sqrt{x^2+y^2}}\right),&(x,y)\neq 0\\0,&(x,y)=0\end{cases}.$$ Then  $f $ is differentiable everywhere but $\dfrac{\partial f}{\partial x}(x,y)$ and $\dfrac{\partial f}{\partial y}(x,y)$ are not continuous at $(0,0)$.

A detailed calculation for the above example can be found in the book Functions of Several Real Variables.
