Answer
$u_{max}= \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt$
and $u_{min}= \lt -\dfrac{1}{3 \sqrt 3}, \dfrac{5}{3 \sqrt 3}, \dfrac{1}{3 \sqrt 3} \gt$
$D_u f_{max} =3 \sqrt 3$
and $D_u f_{min} =-3 \sqrt 3$
Work Step by Step
$ f_x= y^{-1} \implies f_x(4,1,1)= 1$ and $ f_y= \dfrac{-x}{y^2}-z \implies f_y(4,1,1)=-5$ and $ f_z= -y \implies f_z(4,1,1)=-1$
The directions of the unit vector are as follows:
$u_{max}=\dfrac{\nabla f (4,1,1)}{|\nabla f (4,1,1) |}= \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt$
and $u_{min}=- u_{max}= \lt -\dfrac{1}{3 \sqrt 3}, \dfrac{5}{3 \sqrt 3}, \dfrac{1}{3 \sqrt 3} \gt$
We know that $D_u f = \nabla f \cdot u$
Therefore, $D_u f_{max} =\nabla f (4,1,1) \cdot u_{max} =\lt 1,-5,1 \gt \cdot \lt \dfrac{1}{3 \sqrt 3}, - \dfrac{5}{3 \sqrt 3}, - \dfrac{1}{3 \sqrt 3} \gt = \dfrac{1}{3 \sqrt 3}+\dfrac{25}{3 \sqrt 3}+\dfrac{1}{3 \sqrt 3}=3 \sqrt 3$
and $D_u f_{min} =-D_u f_{max} =-3 \sqrt 3$
