= Solution
Set $D=1+x^2+y^2>0$. The <quotient rule> gives the <partial derivatives>
$$
f_x=-\frac{4x(1+y+y^2)}{D^2},\qquad
f_y=\frac{2(1+x^2-y^2+2x^2y)}{D^2}.
$$
Since $1+y+y^2=(y+1/2)^2+3/4>0$, the first equation $f_x=0$ forces $x=0$. The second then reduces to $1-y^2=0$. Thus the only <stationary points> and their field values are
$$
\boxed{(x,y)=(0,1),\ f=1;\qquad (x,y)=(0,-1),\ f=-1.}
$$
No division by a potentially vanishing factor of $y$ was used, so no stationary points have been discarded.
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