Answer
$$\nabla g\left( {2,0} \right) = 2{\bf{i}} + 2{\bf{j}}$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = 2x{e^{y/x}},{\text{ }}\left( {2,0} \right) \cr
& {\text{Calculate the partial derivatives }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right) \cr
& {\text{ }}{g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x{e^{y/x}}} \right] \cr
& {\text{ }}{g_x}\left( {x,y} \right) = 2x\left( { - \frac{y}{{{x^2}}}} \right){e^{y/x}} + {e^{y/x}}\left( 2 \right) \cr
& {\text{ }}{g_x}\left( {x,y} \right) = - 2y{e^{y/x}} + 2{e^{y/x}} \cr
& and \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x{e^{y/x}}} \right] \cr
& {g_y}\left( {x,y} \right) = 2x\left( {\frac{1}{x}} \right){e^{y/x}} \cr
& {g_y}\left( {x,y} \right) = 2{e^{y/x}} \cr
& \cr
& {\text{The gradient of }}g\left( {x,y} \right){\text{ is}} \cr
& \nabla g\left( {x,y} \right) = {g_x}\left( {x,y} \right){\bf{i}} + {g_y}\left( {x,y} \right){\bf{j}} \cr
& \nabla g\left( {x,y} \right) = \left[ { - 2y{e^{y/x}} + 2{e^{y/x}}} \right]{\bf{i}} + 2{e^{y/x}}{\bf{j}} \cr
& {\text{At the point }}\left( {2,0} \right){\text{ the gradient is}} \cr
& \nabla g\left( {2,0} \right) = \left[ { - 2\left( 0 \right){e^{0/2}} + 2{e^{0/2}}} \right]{\bf{i}} + 2{e^{0/2}}{\bf{j}} \cr
& \nabla g\left( {2,0} \right) = 2{\bf{i}} + 2{\bf{j}} \cr} $$
