Answer
$${\text{Saddle point at }}\left( { - \frac{3}{4},\frac{5}{4}, - \frac{{23}}{8}} \right)$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = - {x^2} - 2xy + {y^2} + x - 4y \cr
& {\text{Calculate the first partial derivatives of }}g\left( {x,y} \right) \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - {x^2} - 2xy + {y^2} + x - 4y} \right] \cr
& {g_x}\left( {x,y} \right) = - 2x - 2y + 1 \cr
& and \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - {x^2} - 2xy + {y^2} + x - 4y} \right] \cr
& {g_y}\left( {x,y} \right) = - 2x + 2y - 4 \cr
& \cr
& {\text{Setting both first partial derivatives equal to zero, we have}} \cr
& {g_x}\left( {x,y} \right) = 0,{\text{ }}{g_y}\left( {x,y} \right) = 0 \cr
& - 2x - 2y + 1 = 0{\text{ }}\left( {\bf{1}} \right),{\text{ }} - 2x + 2y - 4 = 0{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Solving the equations simultaneously, we obtain}} \cr
& x = - \frac{3}{4},{\text{ }}y = \frac{5}{4} \cr
& {\text{The critical point is }}\left( { - \frac{3}{4},\frac{5}{4}} \right) \cr
& \cr
& {\text{Find the second partial derivatives of }}g\left( {x,y} \right){\text{ and }}{g_{xy}}\left( {x,y} \right) \cr
& {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 2x - 2y + 1} \right] = - 2 \cr
& {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2x + 2y - 4} \right] = 2 \cr
& {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2x - 2y + 1} \right] = - 2 \cr
& {\text{By the second partials test}} \cr
& d\left( {x,y} \right) = {g_{xx}}\left( {a,b} \right){g_{yy}}\left( {a,b} \right) - {\left[ {{g_{xy}}\left( {x,y} \right)} \right]^2} \cr
& {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( { - \frac{3}{4},\frac{5}{4}} \right) \cr
& d\left( { - \frac{3}{4},\frac{5}{4}} \right) = {g_{xx}}\left( { - \frac{3}{4},\frac{5}{4}} \right){g_{yy}}\left( { - \frac{3}{4},\frac{5}{4}} \right) - {\left[ {{g_{xy}}\left( { - \frac{3}{4},\frac{5}{4}} \right)} \right]^2} \cr
& d\left( { - \frac{3}{4},\frac{5}{4}} \right) = \left( { - 2} \right)\left( 2 \right) - {\left[ { - 2} \right]^2} \cr
& d\left( { - \frac{3}{4},\frac{5}{4}} \right) = - 8 \cr
& d < 0,{\text{ then}} \cr
& g\left( {x,y} \right){\text{ has a saddle point at }}\left( { - \frac{3}{4},\frac{5}{4},g\left( { - \frac{3}{4},\frac{5}{4}} \right)} \right) \cr
& g\left( {x,y} \right) = - {x^2} - 2xy + {y^2} + x - 4y \cr
& g\left( { - \frac{3}{4},\frac{5}{4}} \right) = - {\left( { - \frac{3}{4}} \right)^2} - 2\left( { - \frac{3}{4}} \right)\left( {\frac{5}{4}} \right) + {\left( {\frac{5}{4}} \right)^2} + \left( { - \frac{3}{4}} \right) - 4\left( {\frac{5}{4}} \right) \cr
& g\left( { - \frac{3}{4},\frac{5}{4}} \right) = - \frac{{23}}{8} \cr
& {\text{Saddle point at }}\left( { - \frac{3}{4},\frac{5}{4}, - \frac{{23}}{8}} \right) \cr} $$
