Answer
$$\left( {\frac{2}{3},4} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {\left( {3x - 2} \right)^2} + {\left( {y - 4} \right)^2} \cr
& {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {3x - 2} \right)}^2} + {{\left( {y - 4} \right)}^2}} \right] \cr
& {f_x}\left( {x,y} \right) = 2\left( {3x - 2} \right)\left( 3 \right) \cr
& {f_x}\left( {x,y} \right) = 18x - 12 \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {3x - 2} \right)}^2} + {{\left( {y - 4} \right)}^2}} \right] \cr
& {f_y}\left( {x,y} \right) = 2\left( {y - 4} \right) \cr
& \cr
& {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr
& {f_x}\left( {x,y} \right) = 0 \cr
& 18x - 12 = 0 \cr
& 18x = 12 \cr
& x = \frac{2}{3} \cr
& and \cr
& {f_y}\left( {x,y} \right) = 0 \cr
& 2\left( {y - 4} \right) = 0 \cr
& y = 4 \cr
& {\text{The critical point occurs at}} \cr
& \left( {\frac{2}{3},4} \right) \cr} $$
