Answer
$$\nabla {\text{ }}f\left( {x,y} \right) = {e^{2xy}}\left\langle {2xy + 1,2{x^2}} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {1,0} \right) = \left\langle {1,2} \right\rangle $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = x{e^{2xy}};\,\,\,\,\,\,P\left( {1,0} \right) \cr
& {\text{compute the gradient of }}f\left( {x,y} \right),{\text{ }} \cr
& {\text{first find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right).{\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x{e^{2xy}}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = x\frac{\partial }{{\partial x}}\left[ {{e^{2xy}}} \right] + {e^{2xy}}\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {f_x}\left( {x,y} \right) = x\left( {2y{e^{2xy}}} \right) + {e^{2xy}}\left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = 2xy{e^{2xy}} + {e^{2xy}} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x{e^{2xy}}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = 2{x^2}{e^{2xy}} \cr
& {\text{using }}\nabla {\text{ }}f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr
& {\text{derivatives}} \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left( {2xy{e^{2xy}} + {e^{2xy}}} \right){\bf{i}} + \left( {2{x^2}{e^{2xy}}} \right){\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {2xy{e^{2xy}} + {e^{2xy}},2{x^2}{e^{2xy}}} \right\rangle \cr
& \nabla {\text{ }}f\left( {x,y} \right) = {e^{2xy}}\left\langle {2xy + 1,2{x^2}} \right\rangle \cr
& {\text{evaluate }}\nabla {\text{ }}f\left( {x,y} \right){\text{ at the point }}P \cr
& \nabla {\text{ }}f\left( {1,0} \right) = {e^{2xy}}\left\langle {2xy + 1,2{x^2}} \right\rangle \cr
& \nabla {\text{ }}f\left( {1,0} \right) = {e^{2\left( 1 \right)\left( 0 \right)}}\left\langle {2\left( 1 \right)\left( 0 \right) + 1,2{{\left( 1 \right)}^2}} \right\rangle \cr
& \nabla {\text{ }}f\left( {1,0} \right) = \left\langle {1,2} \right\rangle \cr
& \cr
& \nabla {\text{ }}f\left( {x,y} \right) = {e^{2xy}}\left\langle {2xy + 1,2{x^2}} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {1,0} \right) = \left\langle {1,2} \right\rangle \cr} $$
