= Solution
Write $x_1=t$, $x_2=t^2$. The two inactive constraints are strict exactly for $-2<t<2$. <Complementary slackness> gives $\lambda_1=\lambda_2=0$. The <KKT conditions> reduce to
$$
4t^2-2t\lambda_3=0,\qquad t+2t^2+\lambda_3=0,\qquad\lambda_3\geq0.
$$
If $t=0$ these give $\lambda_3=0$. Otherwise $\lambda_3=2t$ and $t=-3/2$, which gives a negative multiplier and is inadmissible. Therefore
$$
\boxed{x=(0,0),\qquad\lambda=(0,0,0)}
$$
is the only <KKT point> with this active set. Its active <gradient> $(0,1)$ is nonzero, so the <linear independence constraint qualification>, and hence the <Mangasarian–Fromovitz constraint qualification>, holds.
It is \b[not a local maximizer]: $(0,\varepsilon)$ is feasible for small $\varepsilon>0$ and has objective $\varepsilon^2>0=F(0,0)$. The feasible boundary path $(t,t^2)$ for small negative $t$ has objective $2t^3+t^4<0$, so it is not a local minimizer either.
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