Answer
$\dfrac{-3}{2\sqrt {13}}$
Work Step by Step
Our aim is to take the first partial derivative of the given function $h(x,y)$ with respect to $x$, by treating $y$ as constant, and vice versa:
$h_x(1,1)=\dfrac{-y/x^2}{1+\dfrac{y^2}{x^2}}+\sqrt 3 [\dfrac{(y/2)}{\sqrt {1-(\dfrac{xy}{2})^2}}]=\dfrac{1}{2} \\ h_y (1,1)=\dfrac{\dfrac{1}{x}}{1+\dfrac{y^2}{x^2}}+\sqrt 3 [\dfrac{(x/2)}{\sqrt {1-(\dfrac{xy}{2})^2}}]=\dfrac{3}{2} $
Now, the gradient equation is:
$\nabla f (1,-1)= \lt h_x,h_y \gt = \lt \dfrac{1}{2} , \dfrac{3}{2} \gt$
and the directional derivative at that direction is
$D_u h=\lt \dfrac{1}{2} , \dfrac{3}{2} \gt \times \dfrac{\lt 3,-2 \gt }{\sqrt {(3)^2+(-2)^2}} =\dfrac{-3}{2\sqrt {13}}$
