Answer
$$z = 4x + 2y - 3{\text{ and }}z = 4y - 4$$
Work Step by Step
$$\eqalign{
& z = 2{x^2} + {y^2};\,\,\,\left( {1,1,3} \right){\text{ and }}\left( {0,2,4} \right) \cr
& f\left( {x,y} \right) = 2{x^2} + {y^2} \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = 4x \cr
& {f_y}\left( {x,y} \right) = 2y \cr
& {\text{Evaluate at the point }}\left( {1,1,3} \right){\text{ and }}\left( {0,2,4} \right) \cr
& {\text{at }}{f_x}\left( {x,y} \right),\,{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {1,1} \right) = 4\left( 1 \right) = 4 \cr
& {f_y}\left( {1,1} \right) = 2\left( 1 \right) = 2 \cr
& and \cr
& {f_x}\left( {0,2} \right) = 4\left( 0 \right) = 0 \cr
& {f_y}\left( {0,2} \right) = 2\left( 2 \right) = 4 \cr
& \cr
& {\text{An equation of the plane tangent to the surface is}} \cr
& z = f\left( {x,y} \right){\text{ at the point }}\left( {a,b,f\left( {a,b} \right)} \right){\text{ is}} \cr
& {\text{For the point }}\left( {1,1,3} \right) \cr
& z = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_x}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr
& z = 4\left( {x - 1} \right) + 2\left( {y - 1} \right) + 3 \cr
& z = 4x - 4 + 2y - 2 + 3 \cr
& z = 4x + 2y - 3 \cr
& {\text{For the point }}\left( {0,2,4} \right) \cr
& z = 0\left( {x - 0} \right) + 4\left( {y - 2} \right) + 4 \cr
& z = 4\left( {y - 2} \right) + 4 \cr
& z = 4y - 4 \cr
& \cr
& z = 4x + 2y - 3{\text{ and }}z = 4y - 4 \cr} $$
