For $f: \mathbb R^3 \to \mathbb R$, defined by:
$f(x,y,z)=\frac{x^2|y|}{x^2|y|+(x-y)^2+z^2}$ (for $\vec v\neq \vec 0$, and $f(\vec 0)=0$)
I need to find all directions in which exists $\frac{\partial f}{\partial \vec v}(\vec 0)$
The directional derivative is defined as (if exists. I'll address the right-side derivative first, for conveniance - so the absolute value won't complicate it too much):
$\frac{\partial f}{\partial \vec v}(\vec 0)=\lim_{t\to0} \frac{f(t\cdot\vec v)-f(\vec 0)}{t} = \lim_{t\to0^{+}} \frac{1}{t}\cdot\frac{t^3x^2|y|}{t^3x^2|y|+t^2(x-y)^2+t^2z^2}= \lim_{t\to0^{+}}\frac{x^2|y|}{tx^2|y|+(x-y)^2+z^2}=\frac{x^2|y|}{(x-y)^2+z^2}=\frac{x^2|y|}{||\vec v||^2-2xy}=\frac{x^2|y|}{1-2xy}\iff x,y\neq\frac{\pm 1}{\sqrt 2}$
So in any direction in the 3D ball other than the lines spanned by $\lbrace\frac{\pm 1}{\sqrt 2}\left( \matrix{1 \newline 1 \newline 0} \right)\rbrace$
Am I right? I'm not at all sure about that... but if so, I guess the case for the left-side derivative is simillar.
You have \begin{align} \frac{f(t\cdot\vec v)-f(\vec 0)}{t} &= \frac{1}{t}\cdot\frac{t^2|t|\,x^2|y|}{t^2|t|\,x^2|y|+t^2(x-y)^2+t^2z^2}\\ \ \\ &= \frac{1}{t}\cdot\frac{|t|^3\,x^2|y|}{|t|^3\,x^2|y|+|t|^2(x-y)^2+|t|^2z^2}\\ \ \\ &= \frac{|t|}t\,\frac{x^2|y|}{|t|x^2|y|+(x-y)^2+z^2}. \end{align} The second fraction converges to $\frac{x^2|y|}{(x-y)^2+z^2}$, but the first one will go to $1$ or $-1$ depending on whether $t>0$ or $t<0$. So the only way for the derivative to exist is that the second fraction is zero: that is $x^2|y|=0$, so either $x=0$ or $y=0$. We also need to account for the case where the denominator doesn't exist, which is when $x=y$ and $z=0$. So the allowed directions (without normalizing) are $$ \begin{bmatrix} 0\\ u\\ v \end{bmatrix} , \ \ \begin{bmatrix} w\\ 0\\ z \end{bmatrix}. $$ Requiring that the above vectors are nonzero already accounts for the denominator case.