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