Let ${ f(x,y) = \frac{x}{ \left| y \right| } \sqrt{ x^{2} + y^{2} } }$ for $y \ne 0$ and $f(x,y) = 0$ for $y=0$. Does it differentiable in $(0,0)$?
I was able to prove that the function has directional derivative in any direction at $(0,0)$, but I know it doesn't mean that the function is differentiable at this point... I tried to check by differential definition at the point:
${ \lim_{(x,y)\to(x_0,y_0)} \frac{f(x,y)-f(x_0,y_0) - A \triangle x - B \triangle y}{ \sqrt{ {(\triangle x)}^{2} + {(\triangle y)}^{2}} } \iff 0}$
But I got:
${ \lim_{(x,y)\to(0,0)} \frac{f(x,y)-f(0,0) - A \triangle x - B \triangle y}{ \sqrt{ {(\triangle x)}^{2} + {(\triangle y)}^{2}} } = ... = \lim_{(x,y)\to(0,0)} = \frac{x}{|y|} }$
How should I continue from this point?
Your function is not continuous in zero.
Consider a path $x=t$ and $y=t^3$ for $t>0$. Your function becomes $$ \frac{t}{t^3}\sqrt {t^2+t^6} = \frac 1t \sqrt {1+t^4}. $$
Clearly, for $t\to0$ the above expression goes to $+\infty$.
On the other hand, if you consider a path $x=y=t$, your function would go to $0$ as $t\to0$, hence $f$ is not continuous (and therefore, not differentiable) in zero.