Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 21

$0$

As we know that, $u=\dfrac{v}{|v|}$ Then, $u=\dfrac{\lt 2,2,-2 \gt}{\sqrt{2^2+2^2+(-2)^2}}=\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt$ Now $dg=(\nabla g \cdot u) ds=[(\lt0\gt)(\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt)](0.2)$ Thus, $dg=0$