Check differentiability of $f(x,y)=y\sin\frac{1}{x}$ at $(0,0)$ Given , $f(x,y)=y\sin\frac{1}{x}$ when $x\ne0$ and $f(0,0)=0$ . Investigate differentiability at $(0,0)$ .
I've found that it is continuous at $(0,0)$ and the partial derivatives $f_x=0$  $\forall (x,y)$ and $f_y=\begin{cases}
\sin\frac1x & \text{ if } x\ne0\\ 
1 & \text{ if } x= 0
\end{cases}.$
For differentiability it is sufficient to show that both partial derivatives exist and one of them(confused between "both continuous" Or "one of them") is continuous about some neighbourhood of $(0,0)$ .
Now, I'm stuck. Please help how to think.
EDIT
$f(0,y)=y$
 A: (Edited in response to Did and PrithiviRaj's comments.)
$$
\begin{aligned}
f_x (0,0) &= \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h} \\
&= \lim_{h \to 0} \frac{0\sin\frac1h - 0}{h} = 0
\end{aligned} \\
\begin{aligned}
f_y (0,0) &= \lim_{k \to 0} \frac{f(0,k)-f(0,0)}{k} \\
&= \lim_{k \to 0} \frac{k - 0}{k} = 1
\end{aligned}
$$
In the definition of differentiability, approach the point $(0,0)$ via the curve $y = x^2$ to see that the limit
$$ \lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - 0 \cdot h - 1 \cdot k}{\sqrt{h^2+k^2}} \tag{*} \label1$$
can't be defined.  Take $k = h^2$.
\begin{equation}
  \begin{aligned}
    \frac{f(h,h^2)-f(0,0)-h^2}{\sqrt{h^2+h^4}}
    &= \frac{h^2 (\sin \frac{1}{h} - 1)}{\sqrt{h^2+h^4}} \\
    &= \frac{\sin\frac{1}{h} - 1}{\sqrt{1 + \frac{1}{h^2}}}
  \end{aligned}
  \tag{$h\ne0$}
  \label{df1}
\end{equation}
Observe that the denominator $\sqrt{1+1/h^2} \to 1$ as $h \to 0$, but
the limit of $\sin(1/h)$ is undefined.
  To see this,


*

*take $h_n = 1/(n\pi)$ so that $h_n\to0$ and $\sin(1/h_n) = 0$; but

*take $h'_n = 1/((2n+1/2)\pi)$ so that $h'_n\to0$ and $\sin(1/h'_n) = 1$.


Hence \eqref{1} is undefined, and $f$ is not differentiable at $(0,0)$.
