 Solution

For the Lagrange multiplier equations, use $\mathcal{L}=x^{2}y+2y-\lambda (x^2+y^2-1)$. They give $2x(y-\lambda )=0$, $x^2+2=2\lambda y$, and $x^2+y^2=1$. If $x\ne 0$, then $\lambda =y$, implying $x^2+2=2y^2$ and therefore $y^2=1$, contradicting $x\ne 0$. Thus only $x=0$, $y=\pm 1$ 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 $3(1-y^2)>0$. ThereforeThese constrained extrema are flat to quadratic order along the circle, so a vanishing second derivative alone would not classify them.