Solution
= Solution
The <gradient> vanishes when
$$
f_x=x(1+2y)=0,\qquad f_y=y+x^2-y^2=0.
$$
If $x=0$, the second equation gives $y=0$ or $y=1$. If $y=-1/2$, it gives $x^2=3/4$. These exhaust the possibilities. The <stationary points> and their values are
$$
\boxed{\begin{array}{c|c}
(x,y)&f(x,y)\\\hline
(0,0)&0\\
(0,1)&1/6\\
(\sqrt3/2,-1/2)&1/6\\
(-\sqrt3/2,-1/2)&1/6
\end{array}}.
$$