\(\displaystyle{f{{\left({x},{y}\right)}}}={y}{e}^{{-{x}}}\Rightarrow{{f}_{{x}}{\left({x},{y}\right)}}=-{y}{e}^{{-{x}}}\) and \(\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}={e}^{{-{x}}}\). If u is a unit vector in the direction \(\displaystyle\theta={\frac{{{2}\pi}}{{{3}}}}\), then

\(\displaystyle{D}_{{u}}{f{{\left({0},{4}\right)}}}={{f}_{{x}}{\left({0},{4}\right)}}{\cos{{\left({\frac{{{2}\pi}}{{{3}}}}\right)}}}+{{f}_{{y}}{\left({0},{4}\right)}}{\sin{{\left({\frac{{{2}\pi}}{{{3}}}}\right)}}}=-{4}\times{\frac{{-{1}}}{{{2}}}}+{1}\times{\frac{{\sqrt{{{3}}}}}{{{2}}}}={2}+{\frac{{\sqrt{{{3}}}}}{{{2}}}}\)

\(\displaystyle{D}_{{u}}{f{{\left({0},{4}\right)}}}={{f}_{{x}}{\left({0},{4}\right)}}{\cos{{\left({\frac{{{2}\pi}}{{{3}}}}\right)}}}+{{f}_{{y}}{\left({0},{4}\right)}}{\sin{{\left({\frac{{{2}\pi}}{{{3}}}}\right)}}}=-{4}\times{\frac{{-{1}}}{{{2}}}}+{1}\times{\frac{{\sqrt{{{3}}}}}{{{2}}}}={2}+{\frac{{\sqrt{{{3}}}}}{{{2}}}}\)