Answer
$\pm\left(\frac{1}{\sqrt{1189}}(-33 \mathbf{i}+10 \mathbf{j})\right)=\mathbf{v}$
Work Step by Step
We know that the gradient, $\pm \nabla f(x, y)$, will be normal to the level curve:
$6 x y-y=f_{x}(x, y)$
$-33=f_{x}(2,-3)$
$-x+3 x^{2}=f_{y}(x, y)$
$10=f_{y}(2,-3)$
$\pm(-33 \mathbf{i}+10 \mathbf{j})=\pm \nabla f(2,-3)$
$\mathbf{V}=\pm(-33 \mathbf{i}+10 \mathbf{j})$
The unit vector is:
$\pm\left(\frac{1}{\sqrt{1189}}(-33 \mathbf{i}+10 \mathbf{j})\right)=\mathbf{v}$
