∂f/∂x=4y-4x^3; ∂f/∂y=4x-4y^3. ∂f/∂x=0 and ∂f/∂y=0 at turning points, so 4y=4x^3 and 4x=4y^3, y=x^3 and x=y^3.
x=x^9 by substitution so x(1-x^8)=0. The (real) solutions are 0, 1, -1 leading to 0, 1 -1 for y.
x=y=0, x=y=1, x=y=-1 satisfy these conditions simultaneously. So f(x,y)=0=f(0,0), f(x,y)=2=f(1,1) and f(x,y)=2=f(-1,-1).
The points are (x,y,f)=(0,0,0), (1,1,2), (-1,-1,2).