Answer
$${D_{\bf{u}}}f\left( {0,3} \right) = \frac{{\sqrt 3 }}{{18}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{x}{{{y^2}}};\,\,\,\,P\left( {0,3} \right);\,\,\,\,{\bf{u}} = \left\langle {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y} \right) = \frac{1}{{{y^2}}} \cr
& {f_y}\left( {x,y} \right) = - \frac{{2x}}{{{y^3}}} \cr
& \nabla f\left( {x,y} \right) = \frac{1}{{{y^2}}}{\bf{i}} - \frac{{2x}}{{{y^3}}}{\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {0,3} \right) \cr
& \nabla f\left( {0,3} \right) = \frac{1}{{{{\left( 3 \right)}^2}}}{\bf{i}} - \frac{{2\left( 0 \right)}}{{{{\left( 3 \right)}^3}}}{\bf{j}} \cr
& \nabla f\left( {0,3} \right) = \frac{1}{9}{\bf{i}} \cr
& \nabla f\left( {0,3} \right) = \left\langle {\frac{1}{9},0} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {0,3} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {0,3} \right) = \left\langle {\frac{1}{9},0} \right\rangle \cdot \left\langle {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {0,3} \right) = \frac{{\sqrt 3 }}{{18}} \cr} $$
