Answer
$\dfrac{\pi}{\sqrt 2}$
Work Step by Step
$r=\cos 3t i+\sin 3t j +3t k$
and $v(t)=-3 \sin 3t i+3 \cos 3t j +3t k \implies v(\dfrac{\pi}{3})=-3j+3k$
$f(-1,0, \pi) =yz i+xz j +xy k \implies f(-1,0, \pi) =-\pi j$
We know that $D_u f = \nabla f \cdot u$
So, $\nabla f \cdot u = (- pi j) (\dfrac{-1}{\sqrt 2} j+\dfrac{1}{\sqrt 2} k) =\dfrac{\pi}{\sqrt 2}$