Answer
$$2$$
Work Step by Step
Directional derivative: $D_v h = \nabla h \cdot v$
Now, $$h_x= -y \sin xy +\dfrac{1}{x} \\ h_y= -x \sin xy +ze^{xy} \\ h_z=ye^{yz}+z^{-1}$$
So, $\nabla g (1,0,\dfrac{1}{2}) =\lt 1,\dfrac{1}{2},2 \gt$
Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,2,2 \gt}{\sqrt {9}}=\dfrac{\lt 1,2,2 \gt}{3}=\lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt $
Since, $D_v \ h = \nabla h \cdot v$
or, $=\lt 1,\dfrac{1}{2},2 \gt \cdot \lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt$
or, $ =\dfrac{6}{3}$
or, $=2 $
