Answer
$${D_{\bf{u}}}g\left( { - 1,1} \right) = 2$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = {x^2}{y^3};\,\,\,\,P\left( { - 1,1} \right);\,\,\,\,{\bf{u}} = \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{5}{{13}}} \right)}^2} + {{\left( {\frac{{12}}{{13}}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}g\left( {x,y} \right){\text{ is}} \cr
& {g_x}\left( {x,y} \right) = 2x{y^3} \cr
& {g_y}\left( {x,y} \right) = 3{x^2}{y^2} \cr
& \nabla g\left( {x,y} \right) = 2x{y^3}{\bf{i}} + 3{x^2}{y^2}{\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( { - 1,1} \right) \cr
& \nabla g\left( { - 1,1} \right) = 2\left( { - 1} \right){\left( 1 \right)^3}{\bf{i}} + 3{\left( { - 1} \right)^2}{\left( 1 \right)^2}{\bf{j}} \cr
& \nabla g\left( { - 1,1} \right) = - 2{\bf{i}} + 3{\bf{j}} \cr
& \nabla g\left( { - 1,1} \right) = \left\langle { - 2,3} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}g{\text{ at }}\left( { - 1,1} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right\rangle \cr
& {D_{\bf{u}}}g\left( {a,b} \right) = \nabla g\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}g\left( { - 1,1} \right) = \left\langle { - 2,3} \right\rangle \cdot \left\langle {\frac{5}{{13}},\frac{{12}}{{13}}} \right\rangle \cr
& {D_{\bf{u}}}g\left( { - 1,1} \right) = - \frac{{10}}{{13}} + \frac{{36}}{{13}} \cr
& {D_{\bf{u}}}g\left( { - 1,1} \right) = 2 \cr} $$
