directional derivatives in (0,0) I've this function : $f(x,y)= \dfrac{(1+x^2)x^2y^4}{x^4+2x^2y^4+y^8}$ for $(x,y)\ne (0,0)$ and $0$ for $(x,y)=(0,0)$

It's admits directional derivatives at the origin?
A directional derivative needs a direction. Maybe you travel vertically straight through $(0,0)$, along the line $x=0$. Then your function is constantly $0$, so of course that particular directional derivative exists. 

Otherwise, travel in the direction of the line $y=kx$. Then away from $(0,0)$ your function is $$f(x)=\frac{k^4(1+x^2)x^6}{x^4+2k^4x^6+k^8x^8}=\frac{k^4(1+x^2)x^2}{1+2k^4x^2+k^8x^4}$$ which is either identically $0$ if $k=0$, or otherwise is quadratic in nature as $x\to0$. In both cases, the derivative is $0$ at $x=0$.

So all directional derivatives for this function exist at $(0,0)$, and they are all equal to $0$.



Note though what happens if you approach $(0,0)$ along the nonlinear path $x=y^2$...
A function admits directional derivative at a point if its gradient $\nabla{f}$ exists at that point. The gradient of your function is given by,
$$\nabla{f}=\left(\begin{array}{cc} -\frac{2\, x\, y^4\, \left(2\, x^6 + 3\, x^4 - 2\, y^4 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} &amp; \frac{4\, x^2\, y^3\, \left(x^6 + x^4 + x^2 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} \end{array}\right)$$
which doesn't exist at $(x,y)=(0,0)$, i.e. the origin, since you have $\frac{0}{0}$ which is undefined.