= Solution
For the <Lagrange multiplier> equations, use $\mathcal L=x^2y+2y-\lambda(x^2+y^2-1)$. They give $2x(y-\lambda)=0$, $x^2+2=2\lambda y$, and $x^2+y^2=1$. If $x\ne0$, then $\lambda=y$, implying $x^2+2=2y^2$ and therefore $y^2=1$, contradicting $x\ne0$. Thus only $x=0$, $y=\pm1$ remain.
The circle is <compact> and the function is <continuous>, so global extrema exist. On it, eliminating $x^2$ gives $f=3y-y^3$, which is increasing for $-1<y<1$ since its <derivative> is $3(1-y^2)>0$. Therefore
$$
\boxed{f_{\max}=2\text{ at }(0,1),\qquad f_{\min}=-2\text{ at }(0,-1).}
$$
These constrained extrema are flat to quadratic order along the circle, so a vanishing second derivative alone would not classify them.
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