Answer
$u_{max}= \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt \\ u_{min}= \lt -\dfrac{1}{3 \sqrt 3}, \dfrac{5}{3 \sqrt 3}, \dfrac{1}{3 \sqrt 3} \gt \\ D_u f_{max} =3 \sqrt 3 \\ D_u f_{min} =-3 \sqrt 3$
Work Step by Step
$ f_x (4,1,1)=\dfrac{1}{y}=1 \\ f_y (4,1,1)= \dfrac{-x}{y^2}-z =-5$ and $ f_z (4,1,1)= -y =-1$
Since, $D_u f = \nabla f \cdot u$
$u_{max}=\dfrac{\nabla f (4,1,1)}{|\nabla f (4,1,1) |}= \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt$
Now, $u_{min}=- u_{max}= \lt -\dfrac{1}{3 \sqrt 3}, \dfrac{5}{3 \sqrt 3}, \dfrac{1}{3 \sqrt 3} \gt$
So, $D_u \ f_{max} =\nabla f (4,1,1) \cdot u_{max} \\=\lt 1,-5,1 \gt \cdot \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt \\ = \dfrac{1}{3 \sqrt 3}+\dfrac{25}{3 \sqrt 3}+\dfrac{1}{3 \sqrt 3} \\ =3 \sqrt 3$
and $D_u f_{min} =-D_u f_{max} =-3 \sqrt 3$
