If the partial derivatives are continuous then the function is differentiable in the context of 3 dimensions Context: This question has been bugging me for a while, mainly due to no knowledge of linear algebra and availability of only an ugly book Stewart's Calculus. There is a sufficient condition for a function $f(x,y)$ to be differentiable at $(x_0,y_0)$. According to Stewart, the partial derivatives have to be continuous at $(x_0,y_0)$. (But, I saw in other places that they have to be continuous in an open region. (Anyways clarify this issue, first.)) I need to prove that it is a sufficient condition without the use of linear algebra.

Sidenote: For the definition of differentiability for function of two variables, I would use the precise limit definition instead of Stewart's approximation definition.

Trials: I have proved the theorem for every linear path. It is not that difficult. But, I am certain that it is not sufficient. I do not have any intuition of the problem. Just manipulated the definitions to suck out something significant.

Thank you for your help.
You can construct a function $f: \mathbb{R}^2 \to \mathbb{R}$ where the partial derivatives are continuous at $\vec0$, but $f$ isnt even continuous there. Let $C$ denote the closed circle with radius 1 centered $(0,1)$ and define $f:\mathbb{R}^2 \to \mathbb{R}$ as
$$f(x) = \begin{cases}1,&amp;x\in C \mbox{ or } x_2 \leq 0 \\
0, &amp; \mbox{otherwise}\end{cases}.$$

You can prove that for every vector $v \in \mathbb{R}^2$, there is an $\epsilon&gt;0$ such that $f$ is constant on the line segment $L=\{av: -\epsilon &lt; a&lt; \epsilon\}$. Therefore, all directional derivatives at the origin (including partial derivatives) are 0, and are continuous. However, $f$ isnt even continuous at the origin let alone differentiable.

So, I think you are right in saying that the partial derivatives being continuous only at one point is not sufficient.
You can construct a function $f: \mathbb{R}^2 \to \mathbb{R}$ where the partial derivatives are continuous at $\vec0$, but $f$ isnt even continuous there. Let $C$ denote the closed circle with radius 1 centered $(0,1)$ and define $f:\mathbb{R}^2 \to \mathbb{R}$ as
$$f(x) = \begin{cases}1,&amp;x\in C \mbox{ or } x_2 \leq 0 \\
0, &amp; \mbox{otherwise}\end{cases}.$$

You can prove that for every vector $v \in \mathbb{R}^2$, there is an $\epsilon&gt;0$ such that $f$ is constant on the line segment $L=\{av: -\epsilon &lt; a&lt; \epsilon\}$. Therefore, all directional derivatives at the origin (including partial derivatives) are 0, and are continuous. However, $f$ isnt even continuous at the origin let alone differentiable.

So, I think you are right in saying that the partial derivatives being continuous only at one point is not sufficient.