f(x,y)=x3-2xy+y2+5; fx=3x2-2y; fy=-2x+2y; fxx=6x; fyy=2; fxy=-2=fyx.
At critical points fx=0 and fy=0 at the same point:
3x2-2y=0, -2x+2y=0, so y=x and 3x2-2x=0, x(3x-2)=0⇒x=y=0, x=y=⅔.
f(0,0)=5; f(⅔,⅔)=131/27.
Critical points are (0,0,5) and (⅔,⅔,131/27).
For second derivative test D(x,y)=fxx.fyy-(fxy)2:
D(0,0)=-4<0 (saddle-point), D(⅔,⅔)=8-4>0 and fxx>0 (minimum).
(0,0,5) is a saddle-point; (⅔,⅔,131/27) is a minimum.