Answer
$$ - 6$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} - {y^2};\,\,\,\,\,\,P\left( { - 1, - 3} \right);\,\,\,\,\,\left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \cr
& {\text{Let }}{\bf{u}} = \,\left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( {\frac{3}{5}} \right)}^2} + {{\left( { - \frac{4}{5}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y} \right) = 2x \cr
& {f_y}\left( {x,y} \right) = - 2y \cr
& \nabla f\left( {x,y} \right) = 2x{\bf{i}} - 2y{\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( { - 1, - 3} \right) \cr
& \nabla f\left( { - 1, - 3} \right) = 2\left( { - 1} \right){\bf{i}} - 2\left( { - 3} \right){\bf{j}} \cr
& \nabla f\left( { - 1, - 3} \right) = - 2{\bf{i}} + 6{\bf{j}} \cr
& \nabla f\left( { - 1, - 3} \right) = \left\langle { - 2,6} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( { - 1, - 3} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( { - 1, - 3} \right) = \left\langle { - 2,6} \right\rangle \cdot \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \cr
& {D_{\bf{u}}}f\left( { - 1, - 3} \right) = - \frac{6}{5} - \frac{{24}}{5} \cr
& {D_{\bf{u}}}f\left( { - 1, - 3} \right) = - 6 \cr} $$
