Answer
$${\text{Saddle point at }}\left( { - 2,3,5} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} + 3xy + {y^2} - 5x \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} + 3xy + {y^2} - 5x} \right] \cr
& {f_x}\left( {x,y} \right) = 2x + 3y - 5 \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 3xy + {y^2} - 5x} \right] \cr
& {f_y}\left( {x,y} \right) = 3x + 2y \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 + 3y - 5 = 0,{\text{ }}3x + 2y = 0 \cr
& {\text{Solving the system of equations, we obtain}} \cr
& x = - 2,{\text{ }}y = 3 \cr
& {\text{The critical point is }}\left( { - 2,3} \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 + 3y - 5} \right] = 2 \cr
& {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3x + 2y} \right] = 2 \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + 3y - 5} \right] = 3 \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( { - 2,3} \right) \cr
& d\left( { - 2,3} \right) = {f_{xx}}\left( { - 2,3} \right){f_{yy}}\left( { - 2,3} \right) - {\left[ {{f_{xy}}\left( { - 2,3} \right)} \right]^2} \cr
& d\left( { - 2,3} \right) = \left( 2 \right)\left( 2 \right) - {\left[ 3 \right]^2} \cr
& d\left( { - 2,3} \right) = - 5 \cr
& d < 0,{\text{ then}} \cr
& f\left( {x,y} \right){\text{ has a saddle point at }}\left( { - 2,3,f\left( { - 2,3} \right)} \right) \cr
& f\left( {x,y} \right) = {x^2} + 3xy + {y^2} - 5x \cr
& f\left( { - 2,3} \right) = {\left( { - 2} \right)^2} + 3\left( { - 2} \right)\left( 3 \right) + {\left( 3 \right)^2} - 5\left( { - 2} \right) \cr
& f\left( { - 2,3} \right) = 5 \cr
& {\text{Saddle point at }}\left( { - 2,3,5} \right) \cr} $$
