Directional derivative of a function $f(x,y,z)= xyz.$ Suppose there is a function $f(x,y,z)= xyz$ and we have to find its directional derivative along the velocity vector of the curve $r = \cos(3t)i + \sin(3t)j + 3(t) k$ at $t= \pi/3$. Now i assumed that since $(3\cos(\pi/3), 3\sin(\pi/3),3(\pi/3))$ satisfies the level surface $f(x,y,z)=0$ the gradient of $f$ must be perpendicular to $(dr/dt)$ and hence the directional derivative must be $0$. However my calculations are giving me the answer $\pi/(2)^{1/2}$. 
 A: Given $f : \mathbb{R}^3 \rightarrow \mathbb{R}$ defined as $f(x,y,z)=xyz$, the directional derivatives of $f$ at a point $a=r(t_0) \in \mathbb{R}^3$ along a curve $r(t)$ given by
$$
D_{v} f (a) = v(t_0) \cdot \nabla f (a), \quad v(t) = \dot{r}(t)
$$
By $r(t) =(x(t),y(t),z(t))= (\cos 3t, \sin 3t, 3t)$ and $t_0=\pi/3$ you'll have 
$$ \nabla f(r(t)) = \nabla f (x(t),y(t),z(t)) = (yz,xz,xy)(t) \quad \text{and}  \quad  v = \dot{r}(t) = (-3 \sin 3t,3 \cos 3t,3)$$
so
$$
D_v f (a) = \Big(-3 \sin 3t \cdot y(t)z(t) + 3 \cos 3t \cdot x(t)z(t) + 3 \cdot x(t)y(t) \Big)\Big|_{t_0=\pi/3} = 3\pi
$$
The result will equal to yours if we're using unit vel. vector (devide by $|v|$). In your argument above you seems want to use the fact that $v \cdot \nabla f = 0$  along the level curves. However the curve $r(t)$ is not a level curves. It may satisfies $f(x,y,z)=0$ for $t=\pi/3$ but it may not for other $t$ (e.g $t=\pi/12$). But the reason its not zero is not that because the curve is not a level curve, but because $v$ not orthogonal with $\nabla f$ there. Because even if the curve is not a level curve, the directional derivatives may be zero which is can happen if the velocity vector $v$ perpendicular to $\nabla f$ at the evaluated point. In your case above it's not by calculation. 
