Is $f(x,y)$ differentiable on some neighborhood of $(x_0,y_0) ?$ A function $f:O\to \mathbb{R}$ , $O$ is an open subset in $\mathbb{R}^2$,
all of its first partial derivatives $f_{x}$, $f_{y}$ are defined on $O$. If
we assume both $f_{x}$ and $f_{y}$ are differentiable at $(x_0,y_0)\in O$, is  $f(x,y)$  differentiable on some neighborhood of $(x_0,y_0) ?$
From above conditions,it’s apparent that $f(x,y)$ is continuous on a neighborhood of $(x_0,y_0)$ and differentiable at $(x_0,y_0)$. I don’t think  there exists a neighborhood of $(x_0,y_0)$ such that $f(x,y)$  is differentiable on whole of it. My question is how to find the example which fails to differentiate on any neighborhood of $(x_0,y_0)$ but satisfies above conditions?
 A: Below is an example of a function $g:R^2\rightarrow R$ which satisfies the requests. i.e.:

*

*is differentiable at the origin

*its partial derivatives are differentiable at the origin.

*all of its partial derivatives exists in $R^2$.

*is not differentiable at points $(\frac 1 {\pi k},0)$ where k is a natural number.

To define $g$ we'll first claim:
claim: There exists a function $f:R^2 \rightarrow R$ which satisfies:

*

*f is non differential function at the origin

*The partial derivatives of f exists and bounded in $R^2$
Such $f$ exists. Take for example $f=\cases {\frac {x^2y} {x^2+y^2} & $(x,y)\neq (0,0)$\cr 0 & $(x,y)=(0,0)$} $  or $f=\cases {\frac {x^2y^3} {(x-y)^4+x^2y^2} & $(x,y)\neq (0,0)$\cr 0 & $(x,y)=(0,0)$} $ 
proof that the 2 examples satisfy the 2 conditions:

We can prove directly that the 2 requirements above hold.
or we can prove that the requirements are satisfied in each of the 2 examples by noticing that both $f$ are homogeneous function of degree 1, i.e. they satisfy $f(xt,yt)=tf(x,y)$ for all $t\in \mathbb{R}, (x,y) \in \mathbb{R}^2$.
and use the known theorem:

every  homogeneous function of degree 1 which is differential in the origin is linear. i.e. $f(x,y)=x*f_x(0,0)+y*f_y(0,0)$



Now, clearly both examples are not linear so they are not differntiable in the origin.
Now, to prove that the $f_x,f_y$ exists everywhere we observe that $f(x,0)=f(0,y)=0$, so $f_x(0,0)=f_y(0,0)=0$. also, $f$ is elementary elsewhere so $f_x,f_y$ exists in the whole plane. also $f_x,f_y$ continuous everywhere except in the origin.
Now to prove that the partial derivatives are bounded in the whole plane we can use another known theorem:

the partial derivatives of homogenic function of degree k is a homogenic function of degree k-1



so by this theorem, the partial derivatives $f_x,f_y$ are homogenic function of degree 0 thus satisfying $f_x(x_0 t,y_0 t)=f_x(x_0,y_0)$ for each $t\neq 0$, which means that $f_x,f_y$ are constant on every strait lines  intersecting the origin(except the origin). So if we look at the unit circle in the plane $\mathbb{S}^1=\{p\big|\, |p|=1\}$ we can say that the Image (${f_x}|_{\mathbb{S}^1}$)=Image ($f_x$).
Now, as I've mentioned $f_x$ is continuous everywhere except at the origin, so $f_x$  is bounded on the compact set $\mathbb{S}^1$ by the Extreme value theorem. So we conclude it is bounded on the whole plane. thus $f$ satisfy both requirements

Now, finally we define $g$.
$g(x,y)=\cases {x^4*f(\sin(1/x),y) & $x \neq 0$\cr 0 & $x=0$}$ 
where $f$ is a function that satisfy the 2 requirements above.
First we prove that g is differential at the origin.

I assumed that $f$ has bounded partial derivatives (condition 2). Thus, it is also  continuous by the following easy-to-prove theorem:

Given  $f:R^2\rightarrow R$ s.t. its  partial derivatives are bounded in the plane. Then $f$ is continuous everywhere.



So, by the Extreme value theorem $|f(x,y)|<k$ for every $(x,y),\, (x,y)\in [-1,1]\times[-1,1]$. Now, since $-1\leq\sin(1/x)\leq 1$ we conclude that $|f(\sin(1/x),y)|<k$ for every $(x,y),\, x\neq 0,\, |y|\leq 1$. 
So, we conclude that $|g(x,y)|\leq x^4*k$ for $|y|\leq 1$ which yields (using the definition of differentiability) that $g$ is differentiable at the origin.

The rest of the proof is also not hard.
