Is there always a direction in which the directional derivative of a function is zero? Simple question like the title asked, Is there always a direction in which the directional derivative of a function is zero? At origin, wouldn't directional derivative always be 0?
 A: Nice functions?: Yes
If a function $f$ is nice enough that it is differentiable at a point $\mathbf{x}$, the directional derivative in the direction of $\mathbf{v}$ (at least if $\mathbf{v}$ is a unit vector), is given by the dot product of the gradient and $\mathbf{v}$: $\nabla_{\mathbf{v}}f(\mathbf{x})=\nabla f(\mathbf{x})\bullet \mathbf{v}$. If you are working in at least two dimensions, you can do as Hans Engler suggested and take $\mathbf{v}$ in a direction perpendicular to $\nabla f(\mathbf{x})$ to force $\nabla f(\mathbf{x})\bullet \mathbf{v}=0$.
So for differentiable functions, the answer is yes.
All functions?: No
However, in general, the answer is no.
Example
Consider the function $$f\left(x,y\right)=\begin{cases}\dfrac{xy^{2}}{x^{2}+y^{4}}+x+y & \text{ where defined }\\0 & \text{ at }(0,0)\end{cases}$$
This function is not differentiable at $(0,0)$, but all directional derivatives exist at that point. It turns out that none of those directions give a directional derivative of $0$.
Argument
The derivative in the direction of $\langle\pm1,0\rangle$ is $\displaystyle{\lim_{h\to0}}\dfrac{f(\pm h,0)-f(0,0)}{h}=\displaystyle{\lim_{h\to0}}\dfrac{\pm h}{h}=\pm 1$.
Now consider all other directions, unit vectors $\left\langle a,b\right\rangle$, where $a\ne0$. The directional derivative is ${\displaystyle \lim_{h\to0}}\dfrac{f\left(h\left\langle a,b\right\rangle \right)-f\left(0,0\right)}{h}={\displaystyle \lim_{h\to0}}\dfrac{ab^{2}}{a^{2}+h^{2}b^{4}}+a+b=\dfrac{b^{2}}{a}+a+b=\dfrac{a^2+ab+b^2}{a}$. If $ab=0$ then since $a\ne0$ we have $b=0$, so this directional derivative is $a=\pm1$. If $ab>0$ then $a^2+ab+b^2>0$ and if $ab<0$ then $a^2+ab+b^2=(a+b)^2-ab>0$. In all cases, the directional derivative can't be $0$.
Pictures
To get a sense for this visually, the graph of $f$ is 
The parabolic crease is why it's not continuous or differentiable at $(0,0)$.
You can see slices of it at various angles to investigate the directional derivatives:

