Let $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y):= \sqrt{|xy|}$. In order to calculate the directional derivative in the direction $u = (a,b)\in\mathbb{R}^2$ I've made the following inequality (considering $t\neq 0$, $x\neq 0 $ and $y\neq 0$)
$\frac{\sqrt{|(x+ta)(y+tb)|}-\sqrt{|xy|}}{t} = \frac{|(x+ta)(y+tb)| - |xy|}{t[\sqrt{|(x+ta)(y+tb)|}+\sqrt{|xy|}]} = $
$ = \frac{|xy + tbx + tay + t^2 ab| - |xy|}{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} \leq \frac{\big ||xy + tbx + tay + t^2 ab| - |xy|\big|}{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]}\leq\frac{| tbx + tay + t^2 ab| }{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} = \frac{|t|| bx + ay + tab| }{t[\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}]} \leq \frac{| bx + ay + tab| }{\sqrt{|xy + tbx + tay + t^2 ab|}+\sqrt{|xy|}} \to \frac{|ax + by|}{2\sqrt{|xy|}}\text{ as } t\to 0$
It's right that $\operatorname{D}_{u}f(x,y) = \frac{|ax + by|}{2\sqrt{|xy|}}?\;\;\;\;\;\; (*)$ And if it is, what is the other inequality I can use in to prove (*) by squeezing it?