How to show the given function is differentiable...?

Consider the function $f:\mathbb{R^2}\to \mathbb{R}$ defined by$$\begin{equation}f(x,y)=\begin{cases}(x-y)^2\sin \frac{1}{x-y},&\text{ if } x\ne y\\0 ,&\text{ if } x=y\end{cases}\end{equation}$$

Then which of the following is true

(1) f is continuous at $(0,0)$

(2)The partial derivative $f_x$ does not exists at $(0,0)$

(3)$f_x$ continuous at $(0,0)$

(4)$f$ is differentiable at $(0,0)$

For option 2,

$$\begin{equation}\begin{aligned}f_x &=\lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h}\\&=\lim_{h\to 0}\frac{h^2\sin 1/h-0}{h}\\&=\lim_{h\to 0}h\sin 1/h\\&=0\end{aligned}\end{equation}$$

So similarly $f_y$

$$\begin{equation}\begin{aligned}f_y &=\lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h}\\&=\lim_{h\to 0}\frac{h^2\sin 1/(0-h)-0}{h}\\&=\lim_{h\to 0}-h\sin 1/h\\&=0\end{aligned}\end{equation}$$

So both partial derivatives exists. so option 2 is incorrect.

Since $f_x=0$, can I say it is continuous at $(0,0)$....?

For Option 1, to check continuity at $(0,0)$ I used the continuity definition and it is getting continuous at $(0,0)$

It is given correct options are 1 and 4.

And I am not getting how to show it is differentiable. and also why option 3 is incorrect.

Please let me know

Thanks in advance!!