Solution
= Solution
The first <partial derivatives> factor as
$$
f_x=4x(x^2+2y^2-1),\qquad f_y=4y(2x^2+1).
$$
Both vanish at the origin, verifying that it is a <stationary point>. Since $2x^2+1>0$, every <stationary point> has $y=0$; the first equation then gives $x=0,\pm1$. Thus one requested additional point is $\boxed{(1,0)}$; the other is $\boxed{(-1,0)}$.