Answer
$${\text{Saddle point at }}\left( {1, - 1, - 1} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} - xy - {y^2} - 3x - y \cr
& {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - xy - {y^2} - 3x - y} \right] \cr
& {f_x}\left( {x,y} \right) = 2x - y - 3 \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - xy - {y^2} - 3x - y} \right] \cr
& {f_y}\left( {x,y} \right) = - x - 2y - 1 \cr
& {\text{Setting both first partial derivatives equal to zero, we have}} \cr
& {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr
& 2x - y - 3 = 0,{\text{ }} - x - 2y - 1 = 0 \cr
& {\text{Solving the system of equations, we obtain}} \cr
& x = 1,{\text{ }}y = - 1 \cr
& {\text{The critical point is }}\left( {1, - 1} \right) \cr
& {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x - y - 3} \right] = 2 \cr
& {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - x - 2y - 1} \right] = - 2 \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x - y - 3} \right] = - 1 \cr
& {\text{By the second partials test}} \cr
& d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{xy}}\left( {x,y} \right)} \right]^2} \cr
& {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {1, - 1} \right) \cr
& d\left( {1, - 1} \right) = {f_{xx}}\left( {1, - 1} \right){f_{yy}}\left( {1, - 1} \right) - {\left[ {{f_{xy}}\left( {1, - 1} \right)} \right]^2} \cr
& d\left( {1, - 1} \right) = \left( 2 \right)\left( { - 2} \right) - {\left[ { - 1} \right]^2} \cr
& d\left( {8,16} \right) = - 5 \cr
& d < 0,{\text{ then}} \cr
& f\left( {x,y} \right){\text{ has a saddle point at }}\left( {1, - 1,g\left( {1, - 1} \right)} \right) \cr
& f\left( {x,y} \right) = {x^2} - xy - {y^2} - 3x - y \cr
& f\left( {x,y} \right) = {\left( 1 \right)^2} - \left( 1 \right)\left( { - 1} \right) - {\left( { - 1} \right)^2} - 3\left( 1 \right) - \left( { - 1} \right) \cr
& {\text{Saddle point at }}\left( {1, - 1, - 1} \right) \cr} $$