Answer
$${D_u}f\left( {1,4} \right) = - \frac{2}{{\sqrt 5 }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{1}{4}{y^2} - {x^2},{\text{ }}P\left( {1,4} \right),{\text{ }}{\bf{v}} = 2{\bf{i}} + {\bf{j}} \cr
& {\text{Calculate }}\nabla f\left( {x,y} \right) \cr
& \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr
& {f_x}\left( {x,y} \right) = - 2x \cr
& {f_y}\left( {x,y} \right) = \frac{1}{2}y \cr
& \nabla f\left( {x,y} \right) = - 2x{\bf{i}} + \frac{1}{2}y{\bf{j}} \cr
& {\text{Evaluate }}\nabla f\left( {1,4} \right) \cr
& \nabla f\left( {1,4} \right) = - 2\left( 1 \right){\bf{i}} + \frac{1}{2}\left( 4 \right){\bf{j}} \cr
& \nabla f\left( {1,4} \right) = - 2{\bf{i}} + 2{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {2{\bf{i}} + {\bf{j}}} \right| = \sqrt {4 + 1} = \sqrt 5 \cr
& {\bf{v}}{\text{ is not a unit vector, the unit vector in the direction of }}{\bf{v}}{\text{ is:}} \cr
& {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{2{\bf{i}} + {\bf{j}}}}{{\sqrt 5 }} = \frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}} \cr
& {\bf{v}}{\text{ is a unit vector.}} \cr
& {\text{The directional derivative at }}\left( {1,4} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot {\bf{u}} \cr
& {D_u}f\left( {1,4} \right) = \nabla f\left( {1,4} \right) \cdot \left( {\frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}}} \right) \cr
& {D_u}f\left( {1,4} \right) = \left( { - 2{\bf{i}} + 2{\bf{j}}} \right) \cdot \left( {\frac{2}{{\sqrt 5 }}{\bf{i}} + \frac{1}{{\sqrt 5 }}{\bf{j}}} \right) \cr
& {D_u}f\left( {1,4} \right) = - \frac{4}{{\sqrt 5 }} + \frac{2}{{\sqrt 5 }} \cr
& {D_u}f\left( {1,4} \right) = - \frac{2}{{\sqrt 5 }} \cr} $$
