Answer
$$2$$
Work Step by Step
$g_x= 3e^x \cos y z$ and $g_y= -3ze^x \sin yz$ and $g_z=-3ye^x \sin yz$
and $\nabla g (0,0,0) =\lt 3,0,0 \gt$
Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 2,1,-2 \gt}{\sqrt {9}}=\lt \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \gt $
The directional derivative can be found as: $D_v g = \nabla g \cdot v$
Therefore,$D_v g = \nabla g \cdot v=\lt 3,0,0 \gt \cdot \lt \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \gt =2 $
