= Solution
Here the <constraint> surface $x^4+2y^2=3$ is compact and regular, so a <global minimum> exists and satisfies the <Lagrange multiplier> equations
$$
x^3(6x^2-4\lambda)=0,\qquad y(6y^2-4\lambda)=0.
$$
Keep the zero-coordinate cases. If $y=0$, the two points are $(\pm3^{1/4},0)$, with value $3\sqrt3$. 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 $(\pm1,\pm1)$, all with value $5/2$.
These cases exhaust the stated eight constrained <stationary points>. Comparing their values gives
$$
\boxed{\min F=\frac52,\qquad (x,y)\in\{(1,1),(1,-1),(-1,1),(-1,-1)\}.}
$$
For additional classification, substituting $u=x^2\in[0,\sqrt3]$ gives $F(u)=u^3+\tfrac38(3-u^2)^2$, with
$$
F'(u)=\tfrac32u(u-1)(u+3).
$$
It decreases from $u=0$ to $u=1$ and increases from $u=1$ to $u=\sqrt3$. 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.
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