Continuity of a partial derivative I have the function
$$f(x,y)=\begin{cases}
       x^2ysin(\frac1x) & \text{if $x$ is not 0} \\
       0            & \text{if $x=0$}\end{cases}$$
And I need to find the derivative and the partial derivatives, and see if they are continuos.
I´ve already proved that the function is continuos for all (x,y). I´ve also found the derivatives:
$$\frac {\partial f}{\partial x}= \begin{cases} 
                2xysin(\frac1x)-ycos(\frac1x)  & \text{if $x$ is not 0} \\
                0                              & \text{if $x=0$}\end{cases}$$
$$\frac {\partial f}{\partial y}= \begin{cases} 
                x^2sin(\frac1x)  & \text{if $x$ is not 0} \\
                0                & \text{if $x=0$}\end{cases}$$
And I've also proved the continuity of $\frac {\partial f}{\partial y}$,
but I have yet to prove that $\frac {\partial f}{\partial x}$ is continuos (or not), and find the actual derivate of the function.
 If someone could give me a hand, it would be great.
Thanks!

$\frac{\partial{f}}{\partial{x}}(x,y)$ is indeed continuous at $(0,0)$ because:
$\|\frac{\partial{f}}{\partial{x}}(x,y)-0\|=\|2xysin(\frac{1}{x})-ycos(\frac{1}{x})\|\leq\|2xysin(\frac{1}{x})\|+\|ycos(\frac{1}{x})\|\leq2\|xy\|+\|y\| \leq
2\|(x,y)\|^2+\|(x,y)\|\to0$ as $(x,y)\to(0,0)$
