Answer
The critical point is: $(2,-1)$.
There is a relative maximum at $(2,-1)$
Work Step by Step
$f(x,y)=x^{2}+xy+y^{2}-3x-5$
$f_x(x,y)=2x+y-3=0 $
$f_y(x,y)=x+2y=0 \rightarrow x=-2y$
Use the substitution method to solve the system of equations
$f_x(x,y)=2x+y-3=0 \rightarrow 2(-2y)+y-3=0$
$y=-1 \rightarrow x=2$
The critical point is: $(2,-1)$
To identify any extrema:
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=1$
For (2,-1):
$f_{xx}(2,-1)=2$
$f_{yy}(2,-1)=2$
$f_{xy}(2,-1)=1$
$D=2\times2 - (1)^{2}=3 \gt0$ and $f_{xx}(2,-1)=2\gt0 \rightarrow$ There is a relative maximum at $(2,-1)$