Answer
Vector opposite the gradient: $-\mathbf{i}+3 \mathbf{j}=\mathbf{V}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{10}}(-\mathbf{i}+3 \mathbf{j})$
Rate of change: $-\frac{\sqrt{5}}{8}=-\|\nabla f(3,1)\|$
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y)$
\[
\begin{aligned}
&\frac{y}{\sqrt{\frac{x-y}{x+y}}(x+y)^{2}}=f_{x}(x, y) \\
\frac{1}{8 \sqrt{2}}=f_{x}(3,1) & \\
&-\frac{x}{\sqrt{\frac{x-y}{x+y}}(x+y)^{2}}= f_{y}(x, y) \\
-\frac{3}{8 \sqrt{2}} =f_{y}(3,1) \\
\frac{1}{8 \sqrt{2}}(\mathbf{i}-3 \mathbf{j})=\nabla f(3,1)
\end{aligned}
\]
Vector opposite the gradient: $-\mathbf{i}+3 \mathbf{j}=\mathbf{V}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{10}}(-\mathbf{i}+3 \mathbf{j})$
Rate of change: $-\frac{\sqrt{5}}{8}=-\|\nabla f(3,1)\|$
