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: 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: Let $\nu=(\cos\alpha,\sin\alpha)$ be any vector in the plane $\mathbb{R}^2$. Let us calculate the limit (which will depend on alpha) which is the d.d. with respect to $\nu$ of $f$:
$$\frac{\partial f}{\partial\nu}(0,0)=\lim_{t\to0^+}\frac{f(t\cos\alpha,t\sin\alpha)}{t}=\lim\frac{(1+t^2\cos^2\alpha)t^6\cos^2\alpha\sin^4\alpha}{(t^2\cos^2\alpha+t^4\sin^4\alpha)^2t}.$$
By an asymptotic relationship I can get rid of the $t^2$ term above and of the $t^4$ term below, since they go to 0 faster than the rest. Si the limit above is the following:
$$\lim\frac{t^6\cos^2\alpha\sin^4\alpha}{t^4\cos^4\alpha\cdot t}=\lim t\frac{\cos^2\alpha\sin^4\alpha}{\cos^4\alpha}=0.$$
Remember the limit is for $t\to0$. So not only do all d.d.'s exist, but they are also all equal to $\nabla f(0,0)\cdot\nu$, where $\cdot$ is the scalar product. Is it my impression, or this implies the function is differentiable in the origin? And has the zero function as its differential? Well that is interesting. This means the converse of the total differential theorem is not true, since the theorem states that if the partial derivatives exist in a neighborhood of a point and are continuous at that point then the function is differentiable at that point, and here we have a function with derivatives that exist in the neighborhood but are not continuous at the point and yet is differentiable. Which makes us wonder: is there any necessary and sufficient condition to differentiability?
Update:
No, the function is not differentiable, nor even continuous, at the origin. So having all d.d.s and them being equal to gradient times direction versor doesn't imply differentiability. I wonder how I could have thought that. The function is not continuous because approaching from $y=\sqrt x$ we get the limit of $\frac{(1+x^2)x^2x^2}{(x^2+x^2)^2}=\frac{(1+x^2)x^4}{4x^4}=\frac{1+x^2}{4}$, and that limit, for $x\to0$ is $\frac14\neq0$.
And now the dollar that was previously missing is in place, and you can read this update at last :).
A: 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,
$$\nabla{f}=\left(\begin{array}{cc} -\frac{2\, x\, y^4\, \left(2\, x^6 + 3\, x^4 - 2\, y^4 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} & \frac{4\, x^2\, y^3\, \left(x^6 + x^4 + x^2 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} \end{array}\right)$$
which doesn't exist at $(x,y)=(0,0)$, i.e. the origin, since you have $\frac{0}{0}$ which is undefined.
