Answer
$$\left( {3, - 4} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} - 6x + {y^2} + 8y \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - 6x + {y^2} + 8y} \right] = 2x - 6 \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - 6x + {y^2} + 8y} \right] = 2y + 8 \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& {f_x}\left( {x,y} \right) = 0 \cr
& 2x - 6 = 0 \cr
& 2x = 6 \cr
& x = 3 \cr
& and \cr
& {f_y}\left( {x,y} \right) = 0 \cr
& 2y + 8 = 0 \cr
& 2y = - 8 \cr
& y = - 4 \cr
& {\text{The critical point occurs at}} \cr
& \left( {3, - 4} \right) \cr} $$
