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: $3 \sqrt{2}=\sqrt{18}=|\nabla f(1,1,-1)\|$
Work Step by Step
$f$ Increases more quickly in positive direction of $\nabla f(x, y, z)$
\[
\begin{array}{l}
3 x^{2} z^{2}=f_{x}(x, y, z) \\
3=f_{x}(1,1,-1) \\
3 y^{2} z=f_{y}(x, y, z) \\
-3=f_{y}(1,1,-1) \\
2 x^{3} z+y^{3}+1=f_{z}(x, y, z) \\
0=f_{z}(1,1,-1) \\
\nabla f(1,1,-1)=3(\mathbf{i}-\mathbf{j})
\end{array}
\]
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: $3 \sqrt{2}=\sqrt{18}=|\nabla f(1,1,-1)\|$
