Partial derivatives of $f(x,y)=\sqrt{|xy|}$

Question

$f(x,y)=\sqrt{|xy|}$

First question: How to find $f_x(0,0)$ and $f_y(0,0)$? I have figured out this using definition - Both are $0$.

My next question is: How to show that $f_x(0,0)$ and $f_y(0,0)$ are the only directional derivative that exist at $(0,0)$?

Again, using definition, let the direction be $\mathbf{v}=(u,v)$. $\nabla_\mathbf{v}f(0,0) = \lim_{h\rightarrow 0}(f(hu,hv)-f(0,0))/h = \lim_{h\rightarrow 0}h\sqrt{|uv|} = 0$.

What's wrong with my argument?

Answer 1

You miscomputed something.

Note that, given $\mathbf v=(u,v)$, $$\begin{align} \nabla_\mathbf{v}f(0,0) &= \lim \limits_{h\rightarrow 0}\left(\dfrac{f(hu,hv)-f(0,0)}{h}\right)\\ &= \lim \limits_{h\rightarrow 0}\left[\dfrac {\left(\left|h^2uv\right|\right)^{1/2}}{h}\right]\\ &= \lim \limits_{h\to 0}\left(\dfrac {\color{red}| h\color{red}|\sqrt{|uv|}}{h}\right). \end{align}$$

Can you conclude?