What regularity conditions on partial derivatives are equivalent to differentiability? This question is intended to kind of rekindle this old question, which was apparently very hard didn't receive a satisfactory answer. I'm aware that the hope of a definite answer is quite slim, but I'm still very curious to know:

Suppose $f:U\to \Bbb R$ is a function, $U$ is an open subset of $\Bbb R^n$, and $x_0\in U$, and the partial derivatives $\partial_i|_{x_0}f,\,i=1,\cdots,n$ exist. Then what regularity conditions on $\partial_i|_{x_0}f,\,i=1,\cdots,n$ are a sufficient and necessary condition for $f$ to be differentiable at $x_0$?

The mere existence of $\partial_i|_{x_0}f$ is far from enough. The continuity of all of them is over-sufficient. The weakest sufficient condition AFAIK is this one, i.e. the continuity of all but one of them, which is nevertheless still over-sufficient.
Given that we can easily construct a function differentiable at $x_0$ yet has all its partial derivatives discontinuous at $x_0$, like $$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\dfrac{1}{\sqrt{x^2+y^2}}\right) & \text{ if $(x,y) \ne (0,0)$}\\0 &  \text{ if $(x,y) = (0,0)$}.\end{cases}$$ this problem I believe is intrinsically hard. 
Has any research been done that can shed some light on this complicated problem?
 A: This is not a direct answer to your question, but I searched high and low in Dieudonné's Treatise on Analysis, and he gives the following exercise (in the Banach space setting, but no different): 
Suppose $f\colon\Bbb R^2\to\Bbb R$ is continuous. $f$ is differentiable at $(a,b)$ if and only if


*

*The partial derivatives of $f$ exist at $(a,b)$.

*For every $\epsilon>0$ there is $\delta>0$ so that $|s|,|t|<\delta$ imply
$$|f(a+s,b+t)-f(a+s,b)-f(a,b+t)+f(a,b)|\le \epsilon(|s|+|t|).$$


(He goes on to remark that the latter condition is satisfied if, say, $\dfrac{\partial f}{\partial x}(a,b)$ exists and  there's a neighborhood $V$ of $(a,b)$ so that $\dfrac{\partial f}{\partial y}$ exists on $V$ and is continuous.)
A: I also am not aware of any necessary and sufficient condition on partial derivatives that imply differentiabiliy and the theorem in the link is the best result I know. One thing that it is important to observe is that differentiability at $x_0$ implies that all directional derivatives $\frac{\partial f}{\partial v}(x_0)$ exist and that 
$$ \frac{\partial f}{\partial v}(x_0)=\sum_{i=1}^n \frac{\partial f}{\partial x_{i}} (x_0)v_i $$
for every $v\in\mathbb{R}$. However, this condition is only necessary and not sufficient. However, if $f$ is Lipschitz continuous, then it becomes also sufficient. This is all I know about this. 
