= Solution
Completing the square in the <optimization constraint> gives
$$
\left(x_2-\frac32x_1\right)^2=\frac14x_1^2-1.
$$
The left side is nonnegative, so every feasible point has $x_1^2\geq4$. Equality is feasible precisely when $x_2=3x_1/2$ and $x_1=\pm2$. Thus \b[the global minimum is]
$$
\boxed{\min x_1^2=4,\qquad (x_1,x_2)=(2,3)\text{ or }(-2,-3).}
$$
There is \b[no maximum]: for every $|x_1|\geq2$, the values $x_2=3x_1/2\pm\sqrt{x_1^2/4-1}$ satisfy the <optimization constraint>, allowing $x_1^2$ to be arbitrarily large. This direct argument also establishes that the two stationary points are the only <local minima> on the two branches of the constraint curve.
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