Answer
$$ - \frac{{22\pi }}{{13}}$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = \sin \pi \left( {2x - y} \right);\,\,\,\,\,\,P\left( { - 1, - 1} \right);\,\,\,\,\,\left\langle {\frac{5}{{13}}, - \frac{{12}}{{13}}} \right\rangle \cr
& g\left( {x,y} \right) = \sin \left( {2\pi x - y\pi } \right) \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 }}g\left( {x,y} \right){\text{ is}} \cr
& {g_x}\left( {x,y} \right) = 2\pi \cos \pi \left( {2x - y} \right) \cr
& {g_y}\left( {x,y} \right) = - \pi \cos \pi \left( {2x - y} \right) \cr
& \nabla g\left( {x,y} \right) = 2\pi \cos \pi \left( {2x - y} \right){\bf{i}} - \pi \cos \pi \left( {2x - y} \right){\bf{j}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( { - 1, - 1} \right) \cr
& \nabla g\left( { - 1, - 1} \right) = 2\pi \cos \pi \left( {2\left( { - 1} \right) - \left( { - 1} \right)} \right){\bf{i}} - \pi \cos \pi \left( {2\left( { - 1} \right) - \left( { - 1} \right)} \right){\bf{j}} \cr
& \nabla g\left( { - 1, - 1} \right) = 2\pi \left( { - 1} \right){\bf{i}} - \pi \left( { - 1} \right){\bf{j}} \cr
& \nabla g\left( {2, - 3} \right) = \left\langle { - 2\pi ,\pi } \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{5}{{13}}, - \frac{{12}}{{13}}} \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( {2, - 3} \right) = \left\langle { - 2\pi ,\pi } \right\rangle \cdot \left\langle {\frac{5}{{13}}, - \frac{{12}}{{13}}} \right\rangle \cr
& {D_{\bf{u}}}g\left( {2, - 3} \right) = - \frac{{10\pi }}{{13}}, - \frac{{12\pi }}{{13}} \cr
& {D_{\bf{u}}}g\left( {2, - 3} \right) = - \frac{{22\pi }}{{13}} \cr} $$
