Can a function be of class C1 even if its partial derivatives are not continuous? I know that if all partial derivatives of a function f exist and are continuous then the function is said to be of class C1 (continuously differentiable). 
However, I was not able to find whether this is a necessary or a sufficient condition. What I mean is: Could a function be of class C1 even though its partial derivatives are not continuous?

PS. Can a function be continuous/differentiable even if its partial derivatives are not continuous?
Thank you!
 A: No. A function is of class $C^{1}$ if and only if it has continuous partial derivatives. You can refer to Rudin's book for a proof.
A: I was looking for a specific question when I jumped into this old one, and I think that I can share some thoughts about it.
Here's the definition of $C^1$ function:

Let $A$ be an open set of $\mathbb{R}^n$. The set $C^1(A;\mathbb{R}^m)$, $m\geq 1$, is defined to be the set of all the functions $f:A\rightarrow\mathbb{R}^m$ such that all the partial derivatives $$\frac{\partial f_j}{\partial x_i},\quad i=1,\dots,n\quad j=1,\dots,m$$ exist and are continuous.

Hence you can see that by definition a $C^1$ function cannot have discontinuous partial derivatives. For what regards the second part of the question, a function can be continuous without being even differentiable, let alone being $C^1$. An easy example is the absolute value function, which is continuous but not differentiable at $0$.
The very interesting part is whether there exists or not a function which is differentiable but also has discontinuous partial derivatives. Of course there is a theorem that says that if a function is $C^1$ then it is differentiable, but does the converse hold? The answer is no, and a good counterexample is the function
$$
f(x,y)=
\begin{cases}
(x^2+y^2)\sin\bigg(\frac{1}{\sqrt{x^2+y^2}}\bigg),\,\, \text{if}\,\, (x,y)\ne (0,0)\\
0,\,\,\text{if}\,\, (x,y)=(0,0)
\end{cases}
$$
whose differentiable at the origin with differential equal to the zero function, and whose partial derivatives are such that:
$$
\frac{\partial f}{\partial x}(0,0)=0\qquad\frac{\partial f}{\partial y}(0,0)=0
$$
but they are not continuous at $(0,0)$, as you can verify by computing the generic partial derivatives and the limit for $(x,y)\rightarrow (0,0)$.
