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