Answer
$u_{max}= \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt$
and $u_{min}=\lt -\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{1}{3 } \gt$
$D_u f_{max} =3$
and $D_u f_{min} =-3$
Work Step by Step
$g_x= e^y \implies g_x(1, \ln 2, \dfrac{1}{2})= 2$ and $ g_y= xe^y \implies g_y(1, \ln 2, \dfrac{1}{2})= 2$ and $g_z= 2z \implies g_z(1, \ln 2, \dfrac{1}{2})= 1$
The directions of the unit vector are as follows:
$u_{max}=\dfrac{\nabla g(1, \ln 2, \dfrac{1}{2}) }{|\nabla g (1, \ln 2, \dfrac{1}{2}) |}= \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt$
and $u_{min}=- u_{max}= \lt -\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{1}{3 } \gt$
We know that $D_u f = \nabla f \cdot u$
Therefore, $D_u f_{max} =\nabla g(1, \ln 2, \dfrac{1}{2}) \cdot u_{max} =\lt 2,2,1 \gt \cdot \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt =3$
and $D_u f_{min} =-D_u f_{max} =-3$
