= Solution
The <gradient> is $(-y^2-2x-3x^2,2y(1-x))$. Its second component vanishes only if $y=0$ or $x=1$. The latter gives the impossible first-component equation $-y^2-5=0$. With $y=0$, the first component gives $x=0$ or $x=-2/3$. The <Hessian matrix> is
$$
H_g=\begin{pmatrix}-2-6x&-2y\\-2y&2(1-x)\end{pmatrix}.
$$
At $(0,0)$ it has <eigenvalues> $-2,2$, giving a \b[saddle point]. At $(-2/3,0)$ its <eigenvalues> are $2,10/3$, giving a \b[strict local minimum], of value $-4/27$. These are all real <stationary points>. The minimum is local: for example $g(x,0)\to-\infty$ as $x\to+\infty$.
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