I want to show that $f(x,y) = \sqrt{|xy|}$ is not differentiable at $0$.
So my idea is to show that $g(x,y) = |xy|$ is not differentiable, and then argue that if $f$ were differentiable, then so would $g$ which is the composition of differentiable functions $\cdot^2$ and $g$.
But I'm stuck as to how to do this. In the one variable case, to show that $q(x) = |x|$ is not differentiable, I can calculate the limit $\frac{|x + h| - |x|}{h}$ as $h\to 0^+$ and $h\to 0^-$, show that the two one-sided limits are distinct, and conclude that the limit $\lim_{h\to 0}\frac{|x + h| - |x|}{h}$ does not exist.
The reason this is easier is that I do not have to have in mind the derivative of the function $q$ in order to calculate it.
But in the case of $g(x,y) = |xy|$, to show that $g$ is not differentiable at $0$, I would need to show that there does not exist a linear transformation $\lambda:\mathbb{R}^{2}\to\mathbb{R}$ such that
$\lim_{(h,k)\to (0,0)}\frac{\left||hk| - \lambda(h,k)\right|}{|(h,k)|} = 0$
I thought of assuming that I had such a $\lambda$, and letting $(h,k)\to (0,0)$ along both $(\sqrt{t},\sqrt{t})$ and $(-\sqrt{t},-\sqrt{t})$ as $t\to 0^{+}$, but this didn't seem to go anywhere constructive.