Answer
The critical point is: $(\frac{4}{5},\frac{2}{5})$.
There is a saddle point at $(\frac{4}{5},\frac{2}{5})$
Work Step by Step
$f(x,y)=x^{2}+xy-2x-2y+2$
$f_x(x,y)=2x+y-2=0 $
$f_y(x,y)=x-2y=0 \rightarrow x=2y$
Use the substitution method to solve the system of equations
$f_x(x,y)=2x+y-2=0 \rightarrow 2(2y)+y-2=0$
$y=\frac{2}{5} \rightarrow x=\frac{4}{5}$
The critical point is:$(\frac{4}{5},\frac{2}{5})$
To identify any extrema:
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=-2$
$f_{xy}(x,y)=1$
For $(\frac{4}{5},\frac{2}{5})$:
$f_{xx}(\frac{4}{5},\frac{2}{5})=2$
$f_{yy}(\frac{4}{5},\frac{2}{5})=-2$
$f_{xy}(\frac{4}{5},\frac{2}{5})=1$
$D=2\times(-2) - (1)^{2}=-5 \lt 0 \rightarrow$ There is a saddle point at $(\frac{4}{5},\frac{2}{5})$
