Answer
Vector along the gradient: $-\mathbf{i}+\mathbf{j}=\mathbf{V}$
Unit vector: $\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\mathbf{v}$
Rate of change: $\frac{1}{\sqrt{2}}=\|\nabla f(1,2,-2)\|$
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y, z)$
\[
\begin{array}{l}
\frac{1}{z} =f_{x}(x, y, z)\\
-\frac{1}{2}=f_{x}(1,2,-2) \\
-\frac{2 z}{y^{3}}=f_{y}(x, y, z) \\
\frac{1}{2} =f_{y}(1,2,-2) \\
-\frac{x}{z^{2}}+\frac{1}{y^{2}}=f_{z}(x, y, z) \\
0=f_{z}(1,2,-2)
\end{array}
\]
\[
\frac{1}{2}(-\mathbf{i}+\mathbf{j})=\nabla f(1,2,-2)
\]
Vector along the gradient: $-\mathbf{i}+\mathbf{j}=\mathbf{V}$
Unit vector: $\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\mathbf{v}$
Rate of change: $\frac{1}{\sqrt{2}}=\|\nabla f(1,2,-2)\|$