Answer
Saddle point at $(2,1)$
Work Step by Step
Since, we have $f_x(x,y)=2x-4y=0, f_y(x,y)=-4x+2y+6=0$
After solving the above two equations, we get
$x=2,y=1$
Thus the critical point is: $(2,1)$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=-12$ and $D=-12 \lt 0$
Thus, saddle point at $(2,1)$