Answer
$\pm\left(\frac{1}{\sqrt{17}}(-4 \mathbf{i}+\mathbf{j})\right)=\mathbf{v}$
Work Step by Step
We know that gradient, $\pm \nabla f(x, y)$, will be normal to the level curve
$8 x y=f_{x}(x, y)$
$-16=f_{x}(1,-2)$
$4 x^{2}=f_{y}(x, y)$
$4=f_{y}(1,-2)$
$\pm \nabla f(1,-2)=\pm(-16 \mathbf{i}+4 \mathbf{j})$
$\mathbf{V}=\pm(-4 \mathbf{i}+\mathbf{j})$
Thus, the unit vector is:
$\pm\left(\frac{1}{\sqrt{17}}(-4 \mathbf{i}+\mathbf{j})\right)=\mathbf{v}$
