Answer
$$\left( {1,0} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} + xy - 2x - y + 1 \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + xy - 2x - y + 1} \right] \cr
& {f_x}\left( {x,y} \right) = 2x + y - 2 \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + xy - 2x - y + 1} \right] \cr
& {f_y}\left( {x,y} \right) = x - 1 \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& \left\{ {2x + y - 2 = 0{\text{ and }}x - 1 = 0} \right. \cr
& {\text{From the equation }}x - 1 = 0{\text{ we obtain }}x = 1 \cr
& 2x + y - 2 = 0 \cr
& 2\left( 1 \right) + y - 2 = 0 \cr
& y = 0 \cr
& {\text{The critical point occurs at}} \cr
& \left( {1,0} \right) \cr} $$
