$f(x,y)=\sqrt{|xy|}$
I need to calculate partial derivatives at $(0,0)$, and conclude whether it is differentiable there.
$f_x(x,y)=\lim_{h\to 0}{f(x+h,y)-f(x,y)\over h}=0=f_y(x,y)$, so can I conclude that $f_x(0,0)=0$ and $f$ is differentiable there?
$\frac{\partial f}{\partial x}(0,0)=\lim_{t\to0}\frac{f(0+t,0)-f(0,0)}{t}=\lim_{t\to0}0=0$. Similarly for $y$.
The partials exist and are $0$ but that does not necessarily mean that the function is differentiable. It suggests however that if it were going to be differentiable then the linear map we need is the $0$ map. Let $h=(h_{1},h_{2})$ then:
$\frac{f(0+h)-f(0)-Df(0)h}{\|h\|}=\frac{\sqrt{|h_{1}h_{2}|}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}$
Take $h_{1}=h_{2}=t$ and $t\in\mathbb{R}\setminus\{0\}$ then the above is:
$\frac{|t|}{|t|\sqrt{2}}=\frac{1}{\sqrt{2}}$. This is true for all $t\in\mathbb{R}\setminus\{0\}$. So if we approach $(0,0)$ along the line $y=x$ then the limit does not go to $0$. Hence the function is not differentiable.
