 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 isAt $(0,0)$ it has eigenvalues $-2,2$, giving a saddle point. At $(-2/3,0)$ its eigenvalues are $2,10/3$, giving a 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 $x\to +\infty$.