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 $\hat v(t=\frac{\pi}{6}) \cdot \nabla f(-1,0,\pi) =\frac{\sqrt{2}\pi}{2}$.