= Solution
Let $q=x^2+4y^2$. The objective has the exact <sum of squares> decomposition
$$
x^2+6xy+4y^2+3=3-\frac12q+\frac32(x+2y)^2.
$$
For every feasible point, $q\leq8$, so this is at least $3-8/2=-1$. Equality requires simultaneously $q=8$ and $x+2y=0$. Substitution gives $y^2=1$ and $x=-2y$. Both resulting points are feasible. Therefore \b[the global <minimum> is $-1$, attained exactly at $(2,-1)$ and $(-2,1)$]. The lower bound and its equality conditions prove global attainment of the <minimum>, rather than merely identifying <stationary points> on the boundary.
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