Answer
$$2$$
Work Step by Step
$h_x= -y \sin xy +\dfrac{1}{x}$ and $h_y= -x \sin xy +ze^{xy}$ and $h_z=ye^{yz}+z^{-1}$
and $\nabla g (1,0,\dfrac{1}{2}) =\lt 1,\dfrac{1}{2},2 \gt$
Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,2,2 \gt}{\sqrt {9}}=\lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt $
The directional derivative can be found as: $D_v h = \nabla h \cdot v$
Therefore,
$D_v h = \nabla h \cdot v=\lt 1,\dfrac{1}{2},2 \gt \cdot \lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt =\dfrac{6}{3}=2 $
