Answer
$${D_u}f\left( {0, - 2} \right) = - 1$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = xy,{\text{ }}P\left( {0, - 2} \right),{\text{ }}{\bf{v}} = \frac{1}{2}\left( {{\bf{i}} + \sqrt 3 {\bf{j}}} \right) \cr
& {\text{Calculate }}\nabla f\left( {x,y} \right) \cr
& \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr
& {f_x}\left( {x,y} \right) = y \cr
& {f_y}\left( {x,y} \right) = x \cr
& \nabla f\left( {x,y} \right) = y{\bf{i}} + x{\bf{j}} \cr
& {\text{Evaluate }}\nabla f\left( {0, - 2} \right) \cr
& \nabla f\left( {0, - 2} \right) = - 2{\bf{i}} + \left( 0 \right){\bf{j}} \cr
& \nabla f\left( {0, - 2} \right) = - 2{\bf{i}} + 0{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {\frac{1}{2}\left( {{\bf{i}} + \sqrt 3 {\bf{j}}} \right)} \right| = \sqrt {\frac{1}{4} + \frac{3}{4}} = 1 \cr
& {\bf{v}}{\text{ is a unit vector.}} \cr
& {\text{The directional derivative at }}\left( {0, - 2} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot {\bf{v}} \cr
& {D_u}f\left( {0, - 2} \right) = \nabla f\left( {0, - 2} \right) \cdot \left( {\frac{1}{2}{\bf{i}} + \frac{{\sqrt 3 }}{2}{\bf{j}}} \right) \cr
& {D_u}f\left( {0, - 2} \right) = \left( { - 2{\bf{i}} + 0{\bf{j}}} \right) \cdot \left( {\frac{1}{2}{\bf{i}} + \frac{{\sqrt 3 }}{2}{\bf{j}}} \right) \cr
& {D_u}f\left( {0, - 2} \right) = - 1 + 0 \cr
& {D_u}f\left( {0, - 2} \right) = - 1 \cr} $$
