Answer
$${\text{Saddle point at }}\left( {0,0} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = - 4{x^2} + 8{y^2} - 3 \cr
& {\text{Calculate the first partial derivatives}} \cr
& {f_x}\left( {x,y} \right) = - 8x \cr
& {f_y}\left( {x,y} \right) = 16y \cr
& {\text{Calculate the second partial derivatives and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_{xx}}\left( {x,y} \right) = - 8 \cr
& {f_{yy}}\left( {x,y} \right) = 16 \cr
& {f_{xy}}\left( {x,y} \right) = 0 \cr
& {\text{Set the first partial derivatives to zero,}} \cr
& {f_x}\left( {x,y} \right) = - 8x = 0 \Rightarrow x = 0 \cr
& {f_y}\left( {x,y} \right) = 16y = 0 \Rightarrow y = 0 \cr
& {\text{The critical point is}} \cr
& \left( {x,y} \right) = \left( {0,0} \right) \cr
& {\text{Evaluate the Second Derivative Test}} \cr
& D\left( {x,y} \right) = {f_x}\left( {x,y} \right){f_y}\left( {x,y} \right) - {\left( {{f_{xy}}\left( {x,y} \right)} \right)^2} \cr
& {\text{Evaluate }}D\left( {x,y} \right){\text{ at }}\left( {0,0} \right) \cr
& D\left( {0,0} \right) = {f_x}\left( {0,0} \right){f_y}\left( {0,0} \right) - {\left( {{f_{xy}}\left( {0,0} \right)} \right)^2} \cr
& D\left( {0,0} \right) = \left( { - 8} \right)\left( {16} \right) - {\left( 0 \right)^2} \cr
& D\left( {0,0} \right) = - 128 \cr
& D < 0{\text{ }} \Rightarrow {\text{Saddle point at }}\left( {0,0} \right) \cr} $$
