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