Finding Local Minimizers using KKT points (in a degenerate case)

Interesting question. It indeed does not need Lagrange.

Convert the problem to the following.

$$\begin{split} \min_{x\in \mathbb{R}^2}\ & (x_1^2)^2 -2(x_2+\dfrac14)^2 + \dfrac18 \\ \text{subject to }\ & 0\le x_1^2 \le -x_2^2-x_2 \end{split}$$

It's obvious that letting $x_1=0$ doesn't affect the optimal solution, so the problem is again converted to the following.

$$\begin{split} \min_{x_2\in \mathbb{R}}\ & -2(x_2+\dfrac14)^2 + \dfrac18 \\ \text{subject to }\ & 0 \le -(x_2+\dfrac12)^2 + \dfrac14 \Longleftrightarrow -1\le x_2\le 0 \end{split}$$

Compare the endpoints to $-\dfrac14$ and we know $-1$ is farther, so the solution is $-1$ where $x = (0,-1)$.

Note that $x_1^2+x_2^2+x_2\leq 0$ can be written as $$ x_1^2+(x_2+\frac12)^2\le\frac14. $$ Hence $x_1\in[-\frac12,\frac12],x_2\in[-1,0]$. So \begin{eqnarray} x_1^4 -2x_2^2-x_2&=&x_1^4-2(x_2^2+\frac12x_2)\\ &=&x_1^4-2\bigg[(x_2+\frac14)^2-\frac1{16}\bigg]\\ &\ge&-2(x_2+\frac14)^2+\frac18. \end{eqnarray} Since $x_2\in[-1,0]$, one has $-\frac34\le x_1+\frac14\le\frac14$ and hence $$ (x_2+\frac14)^2\le (-\frac34+1)^2=\frac9{16}. $$ So \begin{eqnarray} x_1^4 -2x_2^2-x_2&=&x_1^4-2(x_2^2+\frac12x_2)\\ &=&x_1^4-2\bigg[(x_2+\frac14)^2-\frac1{16}\bigg]\\ &\ge&-2(x_2+\frac14)^2+\frac18\\ &\ge&-2\times\frac9{16}+\frac18\\ &=&-1 \end{eqnarray} and the equal sign holds if and only if $x_1=0$ and $x_2=-1$.

Hint.

According to the convention

$$ \max_x f(x)\ \ \ \text{s. t.}\ \ g(x) \ge 0 $$

our problem can be stated as

$$ \max_x -(x_2^2-x_2^2-x_2) \ \ \ \text{s. t.}\ \ \ -(x_1^2+x_2^2+x_2) \ge 0 $$

so according to this convention, at the relative solutions $x^*$ the vectors $\nabla g(x^*)$ and $\nabla f(x^*)$ should obey

$\nabla f(x^*) \ge \lambda \nabla g(x^*)$ for $\lambda > 0$

Attached a plot showing the level curves for $f(x)$ inside the feasible region. In red dots the stationary points $x^*$, in red vectors $\nabla g(x^*)$ and in black vectors $\nabla f(x^*)$ As we can see there are four stationary points at the boundary and one interior. There are also two candidates to the solution.