Answer
$$ - 6\sqrt 3 + \frac{{27}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 10 - 3{x^2} + \frac{{{y^4}}}{4};\,\,\,\,\,\,P\left( {2, - 3} \right);\,\,\,\,\,\left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \cr
& {\text{Let }}{\bf{u}} = \,\left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y} \right) = - 6x \cr
& {f_y}\left( {x,y} \right) = {y^3} \cr
& \nabla f\left( {x,y} \right) = - 6x{\bf{i}} + {y^3}{\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {2, - 3} \right) \cr
& \nabla f\left( {2, - 3} \right) = - 6\left( 2 \right){\bf{i}} + {\left( { - 3} \right)^3}{\bf{j}} \cr
& \nabla f\left( {2, - 3} \right) = - 12{\bf{i}} - 27{\bf{j}} \cr
& \nabla f\left( {2, - 3} \right) = \left\langle { - 12, - 27} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {2, - 3} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {2, - 3} \right) = \left\langle { - 12, - 27} \right\rangle \cdot \left\langle {\frac{{\sqrt 3 }}{2}, - \frac{1}{2}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {2, - 3} \right) = - 6\sqrt 3 + \frac{{27}}{2} \cr} $$
