Solution
= Solution
Let $D=1+x^2+y^2$. The <gradient> has components $f_x=[D-2x(x+y)]/D^2$ and $f_y=[D-2y(x+y)]/D^2$. At a <critical point>, subtraction gives $(x-y)(x+y)=0$. The case $x+y=0$ cannot make either numerator zero, so $x=y=t$. The remaining equation is $1+2t^2=4t^2$, giving
$$
\boxed{(x,y)=(1/\sqrt2,1/\sqrt2),\quad f=1/\sqrt2;\qquad(x,y)=(-1/\sqrt2,-1/\sqrt2),\quad f=-1/\sqrt2.}
$$
These are the only stationary points; the denominator is positive everywhere.