 Solution

Here the constraint surface $x^4+2y^2=3$ is compact and regular, so a global minimum exists and satisfies the Lagrange multiplier equationsKeep the zero-coordinate cases. If $y=0$, the two points are $(\pm 3^{1/4},0)$, with value $3\sqrt{3}$. If $x=0$, the two points are $(0,\pm \sqrt{3/2})$, with value $27/8$. If both coordinates are nonzero, the equations give $x^2=y^2=t>0$. The constraint becomes $t^2+2t=3$, whose only positive solution is $t=1$. These are the four points $(\pm 1,\pm 1)$, all with value $5/2$.

These cases exhaust the stated eight constrained stationary points. Comparing their values givesFor additional classification, substituting $u=x^2\in [0,\sqrt{3}]$ gives $F(u)=u^3+\frac{3}{8}(3-u^2{)}^2$, withIt decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt{3}$. Thus the two $y$-axis points are constrained local maxima, the two $x$-axis points are global maxima, and the four mixed-coordinate points are global minima.