= Solution
Use $F=x^6+\tfrac32y^4$ and $G=x^4-2y^2-3$. Every <feasible point> has $x\ne0$, so $\nabla G=(4x^3,-4y)$ cannot vanish. The <Lagrange multiplier> equations are
$$
6x^5=4\lambda x^3,\qquad6y^3=-4\lambda y.
$$
The first gives $\lambda=\tfrac32x^2>0$. If $y\ne0$, the second would give $\lambda=-\tfrac32y^2<0$, a contradiction. Therefore $y=0$, and the <constraint> gives $x^4=3$. The only constrained <stationary points> are $(\pm3^{1/4},0)$.
For a global certificate, feasibility implies $x^4\geq3$, and hence
$$
F=(x^4)^{3/2}+\tfrac32y^4\geq3^{3/2}.
$$
Equality requires $y=0$ and $x^4=3$. Thus
$$
\boxed{\min F=3\sqrt3,\qquad (x,y)=(\pm3^{1/4},0).}
$$
The case split is essential: dividing by $y$ would discard precisely the minimizers.
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