Answer
$4 \sqrt{13}=\|\nabla f(-1,1)\|$
Work Step by Step
$f$ increases most rapidly in the positive direction of $\nabla f(x, y)$
\[
\begin{array}{l}
12 x^{2} y^{2}=f_{x}(x, y) \\
12=f_{x}(-1,1) \\
8 x^{3} y =f_{y}(x, y)\\
-8=f_{y}(-1,1)
\end{array}
\]
\[
12 \mathbf{i}-8 \mathbf{j}=\nabla f(-1,1)
\]
$3 \mathbf{i}-2 \mathbf{j}=\mathbf{V}$
The unit vector is:
$\mathbf{v}=\frac{1}{\sqrt{13}}(3 \mathbf{i}-2 \mathbf{j})$
The rate of change is:
$4 \sqrt{13}=\|\nabla f(-1,1)\|$
