Answer
$$u_{max}= \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt \\ u_{min}=\lt -\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{1}{3 } \gt \\ D_u f_{max} =3 \\ D_u f_{min} =-3$$
Work Step by Step
$g_x(1, \ln 2, \dfrac{1}{2})= e^y =2 \\ g_y (1, \ln 2, \dfrac{1}{2})= xe^y = 2 \\ g_z (1, \ln 2, \dfrac{1}{2})= 2z =(2)(1/2)=1$
$u_{max}=\dfrac{\nabla g(1, \ln 2, \dfrac{1}{2}) }{|\nabla g (1, \ln 2, \dfrac{1}{2}) |}= \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt$
$u_{min}=- u_{max}= \lt -\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{1}{3 } \gt$
Since, $D_u f = \nabla f \cdot u$
Now, $D_u \ f_{max} =\nabla g(1, \ln 2, \dfrac{1}{2}) \cdot u_{max} \\=\lt 2,2,1 \gt \cdot \lt \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3} \gt \\=3$
$D_u f_{min} =-D_u f_{max} =-3$
