Answer
$$\eqalign{
& {\text{Relative maximum at the point }}\left( { - 1, - 1,2} \right) \cr
& {\text{Saddle point at }}\left( {0,0,1} \right) \cr
& {\text{Relative maximum at the point }}\left( {1,1,2} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 2xy - \frac{1}{2}\left( {{x^4} + {y^4}} \right) + 1 \cr
& {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2xy - \frac{1}{2}\left( {{x^4} + {y^4}} \right) + 1} \right] \cr
& {f_x}\left( {x,y} \right) = 2y - \frac{1}{2}\left( {4{x^3}} \right) \cr
& {f_x}\left( {x,y} \right) = 2y - 2{x^3} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2xy - \frac{1}{2}\left( {{x^4} + {y^4}} \right) + 1} \right] \cr
& {f_y}\left( {x,y} \right) = 2x - \frac{1}{2}\left( {4{y^3}} \right) \cr
& {f_y}\left( {x,y} \right) = 2x - 2{y^3} \cr
& {\text{Setting both first partial derivatives equal to zero, we have}} \cr
& {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr
& 2y - 2{x^3} = 0,{\text{ }}2x - 2{y^3} = 0 \cr
& y - {x^3} = 0,{\text{ }}x - {y^3} = 0 \cr
& y = {x^3},{\text{ }}x = {y^3} \cr
& y = {y^9} \cr
& y - {y^9} = 0 \cr
& y\left( {1 - {y^8}} \right) = 0 \cr
& {\text{Solving the system of equations, we obtain}} \cr
& y = 0,{\text{ }}y = - 1,{\text{ }}y = 1 \cr
& x = {y^3} \to y = 0,{\text{ }}x = 0 \cr
& x = {y^3} \to y = - 1,{\text{ }}x = - 1 \cr
& x = {y^3} \to y = 1,{\text{ }}x = 1 \cr
& {\text{The critical point are: }}\left( { - 1, - 1} \right),{\text{ }}\left( {0,0} \right){\text{ and }}\left( {1,1} \right) \cr
& \cr
& {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2y - 2{x^3}} \right] = - 6{x^2} \cr
& {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x - 2{y^3}} \right] = - 6{y^2} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2y - 2{x^3}} \right] = 2 \cr
& {\text{By the second partials test}} \cr
& d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{xy}}\left( {x,y} \right)} \right]^2} \cr
& d\left( {x,y} \right) = \left( { - 6{x^2}} \right)\left( { - 6{y^2}} \right) - \left( 2 \right) \cr
& d\left( {x,y} \right) = 36{x^2}{y^2} - 2 \cr
& \cr
& {\text{*Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( { - 1, - 1} \right) \cr
& d\left( { - 1, - 1} \right) = 36{\left( { - 1} \right)^2}{\left( { - 1} \right)^2} - 2 \cr
& d\left( { - 1, - 1} \right) = 34 > 0 \cr
& d > 0,{\text{ and }}{f_{xx}}\left( { - 1, - 1} \right) = - 6 < 0 \cr
& d > 0{\text{ and }}{f_{xx}}\left( { - 1, - 1} \right) < 0 \cr
& {\text{then}} \cr
& f\left( {x,y} \right){\text{ has a relative maximum at }}\left( { - 1, - 1,f\left( { - 1, - 1} \right)} \right) \cr
& f\left( { - 1, - 1} \right) = 2 \cr
& {\text{Relative maximum at the point }}\left( { - 1, - 1,2} \right) \cr
& \cr
& {\text{*Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {0,0} \right) \cr
& d\left( {0,0} \right) = 36{\left( 0 \right)^2}{\left( 0 \right)^2} - 2 \cr
& d\left( {0,0} \right) = - 2 < 0 \cr
& f\left( {x,y} \right){\text{ has a saddle point at }}\left( {0,0,f\left( {0,0} \right)} \right) \cr
& f\left( {0,0} \right) = 1 \cr
& {\text{Saddle point at }}\left( {0,0,1} \right) \cr
& \cr
& {\text{*Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {1,1} \right) \cr
& d\left( {1,1} \right) = 36{\left( 1 \right)^2}{\left( 1 \right)^2} - 2 \cr
& d\left( {1,1} \right) = 34 > 0 \cr
& d > 0,{\text{ and }}{f_{xx}}\left( {1,1} \right) = - 6 < 0 \cr
& d > 0{\text{ and }}{f_{xx}}\left( {1,1} \right) < 0 \cr
& {\text{then}} \cr
& f\left( {x,y} \right){\text{ has a relative maximum at }}\left( {1,1,f\left( {1,1} \right)} \right) \cr
& f\left( {1,1} \right) = 2 \cr
& {\text{Relative maximum at the point }}\left( {1,1,2} \right) \cr
} $$