$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The second edition
Let $f$ be a function of two variables,let$(x_{0},y_{0})$ be a point and let $U$ be an open disk with center $(x_{0},y_{0})$.Assume that $f$ is defined on $U$ and its mixed partial derivatives $f_{xy}$ ,$f_{yx }$ exist on $U.$
If $f_{xy}$ , $f_{yx}$ are continuous at $(x_{0},y_{0})$,then $f_{x},f_{y}$ are continuous at $(x_{0},y_{0})$?
I think $\frac{\partial^{2} f}{\partial x \partial y},\frac{\partial^{2} f}{\partial y \partial x} $ are contiunous at $(x_{0},y_{0})$ $\nRightarrow$ $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y }$ are continuous at $(x_{0},y_{0})$. I want to find a counterexample to support my argument.
According to Daniel Fischer's best hints,I give my counterexample:
$f(x,y)=\left\{\begin{matrix} x^{2}\sin(\frac{1}{x})+y^{2}\sin(\frac{1}{y})\quad \text{as} \quad xy\ne0\quad; \\ \qquad \qquad x^{2}\sin(\frac{1}{x})\qquad\qquad\quad \text{as} \quad x\ne0 \vee y=0\quad; \\ \qquad \qquad y^{2}\sin(\frac{1}{y})\qquad\qquad\quad \text{as} \quad x=0 \vee y\ne0\quad; \\ \quad\quad\quad\qquad0\qquad\qquad\qquad\quad\text{as} \quad x=0 \vee y=0\quad. \end{matrix}\right.\Longrightarrow $
$\frac{\partial f}{\partial x}(x,y)=\left\{\begin{matrix} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x})& \text{as} \quad x\ne0\\ 0 & \text{as} \quad x=0 \end{matrix}\right.\quad$and$\frac{\partial f}{\partial y}(x,y)=\left\{\begin{matrix} 2y\sin(\frac{1}{y})-\cos(\frac{1}{y})& \text{as} \quad y\ne0\\ 0 & \text{as} \quad y=0 \end{matrix}\right.\quad$.$\Longrightarrow $ $\frac{\partial^{2} f}{\partial x \partial y}(x,y)\equiv0\equiv\frac{\partial^{2} f}{\partial y \partial x}(x,y)!$ So $f(x,y)$ is enough to satisfy my argument.
Anyone can give more simple counterexamples ? Your reply will be higly appreciated!