Find the derivative of $f(x, y, z) = xyz$ in the direction of the velocity vector of the helix
$r(t) = (\cos (6t) , \sin (6t) , 6t)$ at $t = \pi/6$.
I am not really sure how to solve this, since we are not explicitly given a point at which to evaluate $f_x = yz, f_y = xz, \text{ and } f_z = xy$.
On
Given the position vector $\vec r(t)=\hat x \cos(6t)+\hat y \sin(6t)+\hat z 6t$, the velocity $\vec v(t)$ is
$$ \begin{align} \frac{d\vec r(t)}{dt} &= \hat x x'(t)+\hat y y'(t) +\hat z z'(t) \\ & = -6 \sin(6t)\hat x+6 \cos(6t) \hat y +6\hat z \end{align}$$
At $t=\frac{\pi}{6}$, the velocity is $\vec v(\frac{\pi}{6})=-6\hat y+6 \hat z$, for which the unit vector $\hat v$ is $\frac{\sqrt{2}}{2}(-\hat y+\hat z)$.
Now, the gradient at $t=\frac{\pi}{6}$ is the gradient at $x=-1$, $y=0$, and $z= \pi$. Thus, $\nabla f(-1,0,\pi)=-\hat y \pi$.
The directional derivative is then $\hat v(t=\frac{\pi}{6}) \cdot \nabla f(-1,0,\pi) =\frac{\sqrt{2}\pi}{2}$.
I would assume that it wants the position defined by $r(\frac{\pi}{6})$. That would give you the point $x = \cos (\pi) = -1, y = \sin (\pi) = 0, z = \pi$. Then you can find the velocity vector $r'(\frac{\pi}{6})$, normalize it (divide by magnitude) and dot it with the gradient to find the directional derivative: $D_{r(\pi/6)}f = \nabla f \cdot \frac{r'(\pi/6)}{||r'(\pi/6)||}$.