So i have to prove that this functions is not differentiable at $(0,0)$:
$f(x)= \left\lvert x \right\rvert+\left\lvert y \right\rvert$
Seems pretty obvious, but i don't know how to do it formally, i can prove that the partial derivatives don't exist (i obtain different numbers if a aproach from ($h->0+$) or ($h->0-$)) but i don't know if that is enough to show that the function is not differentiable.
i can decompose the function in 4
$ f(x) = \begin{cases} x+y, & \text{if $x$ > 0 & $y$ > 0} \\\ x-y, & \text{if $x$ > 0 & $y$ < 0} \\\ -x+y, & \text{if $x$ < 0 & $y$ > 0} \\\ -x-y, & \text{if $x$ < 0 & $y$ < 0} \end{cases}$
but then i don't know exactly what to do. Show that the function has different tangent planes in each quadrant? Is that enough?
My final question is "Is there a function that is differentiable at a point but it doesn't have partial derivatives at that point?"
If a function is differentiable at a point, then all partial derivatives exist at that point. What fails is the converse: the existence of the partial derivatives is not enough to imply differentiability (continuity of the partial derivatives is).