Answer
$$\nabla f\left( {x,y} \right) = 2\left[ {\left( {x + y} \right){\bf{i}} + x{\bf{j}}} \right],{\text{ }}2\sqrt 2 $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} + 2xy,{\text{ point }}\left( {1,0} \right) \cr
& {\text{Find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + 2xy} \right] \cr
& {f_x}\left( {x,y} \right) = 2x + 2y \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 2xy} \right] \cr
& {f_y}\left( {x,y} \right) = 2x \cr
& {\text{Calculate }}\nabla f\left( {x,y} \right) \cr
& \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr
& \nabla f\left( {x,y} \right) = \left( {2x + 2y} \right){\bf{i}} + 2x{\bf{j}} \cr
& \nabla f\left( {x,y} \right) = 2\left[ {\left( {x + y} \right){\bf{i}} + x{\bf{j}}} \right] \cr
& {\text{At the given point the gradient is}} \cr
& \nabla f\left( {1,0} \right) = \left( {2 + 0} \right){\bf{i}} + 2\left( 1 \right){\bf{j}} \cr
& \nabla f\left( {1,0} \right) = 2{\bf{i}} + 2{\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ is }}\left\| {\nabla f\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla f\left( {1,0} \right)} \right\| = \left\| {2{\bf{i}} + 2{\bf{j}}} \right\| \cr
& \left\| {\nabla f\left( {1,0} \right)} \right\| = \sqrt {{2^2} + {2^2}} \cr
& \left\| {\nabla f\left( {1,0} \right)} \right\| = \sqrt 8 \cr
& \left\| {\nabla f\left( {1,0} \right)} \right\| = 2\sqrt 2 \cr} $$