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