Answer
$$18$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3{x^2} + {y^3};\,\,\,\,\,\,P\left( {3,2} \right);\,\,\,\,\,\left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr
& {\text{Let }}{\bf{u}} = \,\left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{5}{{13}}} \right)}^2} + {{\left( {\frac{{12}}{{13}}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y} \right) = 6x \cr
& {f_y}\left( {x,y} \right) = 3{y^2} \cr
& \nabla f\left( {x,y} \right) = 6x{\bf{i}} + 3{y^2}{\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {3,2} \right) \cr
& \nabla f\left( {3,2} \right) = 6\left( 3 \right){\bf{i}} + 3{\left( 2 \right)^2}{\bf{j}} \cr
& \nabla f\left( {3,2} \right) = 18{\bf{i}} + 12{\bf{j}} \cr
& \nabla f\left( {3,2} \right) = \left\langle {18,12} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {3,2} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {3,2} \right) = \left\langle {18,12} \right\rangle \cdot \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {3,2} \right) = \frac{{90}}{{13}} + \frac{{144}}{{13}} \cr
& {D_{\bf{u}}}f\left( {3,2} \right) = 18 \cr} $$
