Answer
$$\left( {0,2} \right),\,\,\,\left( { - 1,2} \right),\,\,\,\left( {1,2} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^4} - 2{x^2} + {y^2} - 4y + 5 \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^4} - 2{x^2} + {y^2} - 4y + 5} \right] \cr
& {f_x}\left( {x,y} \right) = 4{x^3} - 4x \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^4} - 2{x^2} + {y^2} - 4y + 5} \right] \cr
& {f_y}\left( {x,y} \right) = 2y - 4 \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& 4{x^3} - 4x = 0{\text{ and }}2y - 4 = 0 \cr
& {\text{From the equation }}{x^2} + 3y = 0 \cr
& 4{x^3} - 4x = 0 \cr
& 4x\left( {{x^2} - 1} \right) = 0 \cr
& {x_1} = 0,\,\,\,{x_2} = 1,\,\,\,{x_3} = - 1 \cr
& and \cr
& 2y - 4 = 0 \cr
& y = 2 \cr
& {\text{The critical points occur at}} \cr
& \left( {0,2} \right),\,\,\,\left( { - 1,2} \right),\,\,\,\left( {1,2} \right) \cr} $$
