Non differentiable function at a point with continuous partial derivatives at that point. Is it possible? There is a theorem that says: if a function $f$ has continuous partial derivatives at a point $x$; and existing partial derivatives in a neighborhood of that point $x$; then the function is differentiable in $x$.
Let’s define this function: $f:R^2\rightarrow R$; $f(x,y)=1$ if $xy≠0$, $f(x,y)=0$ if $xy=0$.
$f$ is clearly not continuous at $(0,0)$ (and hence not differentiable), but according to some definitions it’s first partial derivatives are still continuous at $(0,0)$.
This shouldn’t be a problem since it’s partial derivatives are not defined in every point in a neighborhood of $(0,0)$, so the theorem hypothesis are not met.

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*Is there any example of a function similar to this one (meaning it has continuous partial derivatives at a point $x$, but is not differentiable at that point $x$ because it’s partials are not defined in a neighborhood of $x$) but is also continuous at $x$?

Moreover, on Wikipedia they say $C^1(U) \subset C^0(U)$ for an open set $U$.


*Having defined $U=\{(x,y):xy≠0\}\cup(0,0)$; Surely $f \notin C^0(U)$ because it’s not continuous at $(0,0)$, but it’s partial derivatives are still continuous in $U$, so  $f \in C^1(U)$, is this possible only because $U$ is not an open set?

Please correct me if any of my statements are wrong. Thank you
 A: The theorem is usually stated in the form: A $C^1$ function (on an open set) is differentiable; $C^1$ indicates that the function and its partial derivatives are continuous. (A function is $C^k$ if all partial derivatives of order $\le k$ exist and are continuous. This means that $f$ is continuous when we talk about partial derivatives of order $0$.)
If you look at the usual proof of the theorem, it is based on two applications of the one-variable Mean Value Theorem. This, in particular, necessitates knowing that the function is continuous on the appropriate horizontal and vertical line segments. But this is a consequence of the existence of the partial derivatives on those segments. So, in fact, continuous partial derivatives on a neighborhood guarantees horizontal and vertical continuity, and this is enough to prove differentiability. But we know that differentiability implies continuity. Interesting, eh?
A: To answer the direct question, no it is not possible for a function to simultaneously have defined, continuous partial derivatives and to not be differentiable at any given point. Suppose that we were to take some surface defined by a function $z = f(x,y)$, then, in the spirit of Calc 1, we might define an incrementation of $z$ at a point $(x_0,y_0)$ as follows:
$$\delta z = f(x_0 + \delta x,y_0 + \delta y) - f(x_0,y_0).$$
We can then say that a function $f$ is differentiable at a point $(x_0,y_0)$ if the incrementation $\delta z$ can be written as
$$\delta z = f_x(x_0,y_0)\delta x + f_y(x_0,y_0)\delta y + \varepsilon_1 \delta x + \varepsilon_2 \delta y$$
where the $\varepsilon_i\rightarrow 0$ as $(\delta x, \delta y) \rightarrow 0$. From this, it should be pretty immediately clear that, if the partial derivatives are defined and continuous at a point, then the function is necessarily differentiable at that point.
At a higher level, we might say that a function $F:V\rightarrow W$ between finite dimensional vector spaces (e.g. $\mathbb{R}^n$) is differentiable at a point $p\in U\subseteq V$ if there is a linear map $A:V\rightarrow W$ such that
$$\lim_{v\rightarrow 0}\frac{|F(p + v) - F(p) - Av|}{|v|} = 0$$
(note that this is just a suped-up version of the Calc I definition of derivative. Indeed, whenever you try to define a derivative, you almost always use a definition of this form). The linear map $A$ satisfying this relationship is called the total derivative of $F$, commonly denoted $DF$. When we take $V,W=\mathbb{R}^n$ this map is just the Jacobian matrix. For example, if we were to take $$F:\mathbb{R}^2 \rightarrow \mathbb{R}^2:(x,y)\mapsto (F_1(x,y),F_2(x,y)) = (x^2y, \sin(y)),$$ then the total derivative may be given by
$$DF =  \begin{pmatrix}DF_1\\DF_2\end{pmatrix} = \begin{pmatrix}\frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y}\\\frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y}\end{pmatrix} =  \begin{pmatrix}2xy & x^2 \\ 0 & \cos(y)\end{pmatrix}.$$
Now, the question, as posed, only concerns a map $G:\mathbb{R}^2\rightarrow \mathbb{R}$, so we might just take $G = F_1$, but note that regardless of the space we are working in, we defined a function to be differentiable so long as the total derivative for the function exists (and is continuous)! So, as long as the components of $DF$ are defined and continuous near the point $p$, the function is necessarily differentaible.
