Answer
\[\left( {1,y} \right){\text{ and }}\left( {x,0} \right)\,\,\,{\text{Where }}x{\text{ and }}y \in \,\,\mathbb{R}\]
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}} \right] \cr
& {f_x}\left( {x,y} \right) = \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}} \right] \cr
& {f_y}\left( {x,y} \right) = \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& {f_x}\left( {x,y} \right) = \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr
& \left( {2x{y^2} - 2{y^2}} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} = 0 \cr
& {e^{{x^2}{y^2} - 2x{y^2} + {y^2}}}{\text{ is allways positive}}{\text{, then}} \cr
& 2x{y^2} - 2{y^2} = 0 \cr
& 2{y^2}\left( {x - 1} \right) = 0 \cr
& x = 1,\,\,\,\,\,y = 0 \cr
& \cr
& and \cr
& \cr
& {f_y}\left( {x,y} \right) = \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} \cr
& \left( {2{x^2}y - 4xy + 2y} \right){e^{{x^2}{y^2} - 2x{y^2} + {y^2}}} = 0 \cr
& 2{x^2}y - 4xy + 2y = 0 \cr
& \cr
& {\text{We know that }}x = 1{\text{ and }}y = 0 \cr
& {\text{For }}x = 1 \cr
& 2{\left( 1 \right)^2}y - 4\left( 1 \right)y + 2y = 0 \cr
& 0 = 0 \cr
& {\text{Infinite solutions}}{\text{, then}} \cr
& \left( {1,y} \right){\text{ Where }}y{\text{ is any real number}} \cr
& and \cr
& {\text{For }}y = 0 \cr
& 2{x^2}\left( 0 \right) - 4x\left( 0 \right) + 2\left( 0 \right) = 0 \cr
& 0 = 0 \cr
& {\text{Infinite solutions}}{\text{, then}} \cr
& \left( {x,0} \right){\text{ Where }}x{\text{ is any real number}} \cr
& \cr
& {\text{The critical point occurs at}} \cr} $$
\[\left( {1,y} \right){\text{ and }}\left( {x,0} \right)\,\,\,{\text{Where }}x{\text{ and }}y \in \,\,\mathbb{R}\]