Continuity of a partial derivative

Question

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!

Answer 1

Notice that $\sin$ and $\cos$ remain bounded, while the factors they have in front tend to zero as $(x,y)$ tends to $(0,0)$.

But now consider the limit to points $(0,y)$ with $y\neq0$.

Answer 2

$\frac{\partial{f}}{\partial{x}}(x,y)$ 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 $(x,y)\to(0,0)$