Answer
$\nabla f(0,-3,0)=\frac{1}{6}(\mathbf{i}+3 \mathbf{j}+4 \mathbf{k})$
Vector along the gradient: $\mathbf{V}=\mathbf{i}+3 \mathbf{j}+4 \mathbf{k}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{26}}(\mathbf{i}+3 \mathbf{j}+4 \mathbf{k})$
Rate of change: $\sqrt{\frac{13}{18}}=\|\nabla f(0,-3,0)\| $
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y, z)$
\[
\begin{array}{l}
\frac{1}{2 \sqrt{-3 y+4 z+x}}=f_{x}(x, y, z) \\
\frac{1}{6} =f_{x}(0,-3,0)\\
-\frac{3}{2 \sqrt{-3 y+4 z+x}}=f_{y}(x, y, z) \\
\frac{1}{2}=f_{y}(0,-3,0) \\
\frac{2}{\sqrt{4 z+x-3 y}}=f_{z}(x, y, z) \\
\frac{2}{3}=f_{z}(0,-3,0)
\end{array}
\]
$\nabla f(0,-3,0)=\frac{1}{6}(\mathbf{i}+3 \mathbf{j}+4 \mathbf{k})$
Vector along the gradient: $\mathbf{V}=\mathbf{i}+3 \mathbf{j}+4 \mathbf{k}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{26}}(\mathbf{i}+3 \mathbf{j}+4 \mathbf{k})$
Rate of change: $\sqrt{\frac{13}{18}}=\|\nabla f(0,-3,0)\| $
