Answer
$\dfrac{21}{13}$
Work Step by Step
Our aim is to take the first partial derivative of the given function $g(x,y)$ with respect to $x$, by treating $y$ as a constant, and vice versa:
$g_x=\dfrac{(xy+2)(1)-y(x-y)}{(xy+2)^2}=\dfrac{y^2+2}{(xy+2)^2} \\ g_y=\dfrac{(xy+2)(-1)-x (x-y)}{(xy+2)^2}=-\dfrac{x^2+2}{(xy+2)^2} $
The gradient equation is:
$\nabla f (1,-1)= \lt \dfrac{y^2+2}{(xy+2)^2}, -\dfrac{x^2+2}{(xy+2)^2} \gt = \lt \dfrac{(-1)^2+2}{(-1+2)^2}, -\dfrac{1+2}{(-1+2)^2} \gt =\lt 3,-3 \gt$
The directional derivative at that direction is given as: $D_v f=\nabla f \cdot v=\lt 3,-3 \gt \times \dfrac{\lt 12, 5 \gt }{\sqrt {(12)^2+(5)^2}} =\dfrac{21}{13}$
