Answer
$\dfrac{-3}{2\sqrt {13}}$
Work Step by Step
In order to find the partial derivative, we will differentiate
with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa:
$h_x(1,1)=\dfrac{-y/x^2}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(y/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{1}{2} \\ h_y (1,1)=\dfrac{1/x}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(x/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{3}{2} $
Write the gradient equation.
$\nabla f (1,-1)= \lt h_x,h_y \gt = \lt \dfrac{1}{2} , \dfrac{3}{2} \gt$
Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3,-2 \gt }{\sqrt {(3)^2+(-2)^2}} =\lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt$
The directional derivative at that direction is given as:
$D_u h=\lt \dfrac{1}{2} , \dfrac{3}{2} \gt \times \lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt =\dfrac{-3}{2\sqrt {13}}$
