Answer
Vector opposite the gradient: $\mathbf{V}=3 \mathbf{i}-\mathbf{j}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{10}}(3 \mathbf{i}-\mathbf{j})$
Rate of change: $-\sqrt{5}=-\left\|\nabla f\left(\frac{\pi}{6}, \frac{\pi}{4}\right)\right\|$
Work Step by Step
$f$ increases more quickly in the positive direction of $\nabla f(x, y)$
\[
\begin{array}{l}
-3 \sin (-y+3 x)=f_{x}(x, y) \\
f_{x}\left(\frac{\pi}{6}, \frac{\pi}{4}\right)=-\frac{3}{\sqrt{2}} \\
\sin 3 x-y=f_{y}(x, y) \\
\frac{1}{\sqrt{2}}=f_{y}\left(\frac{\pi}{6}, \frac{\pi}{4}\right)
\end{array}
\]
$\nabla f\left(\frac{\pi}{6}, \frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(-3 \mathbf{i}+\mathbf{j})$
Vector opposite the gradient: $\mathbf{V}=3 \mathbf{i}-\mathbf{j}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{10}}(3 \mathbf{i}-\mathbf{j})$
Rate of change: $-\sqrt{5}=-\left\|\nabla f\left(\frac{\pi}{6}, \frac{\pi}{4}\right)\right\|$
