Your question is
Find the minimum of $f(x,y)=(x-2)^2+y$ subject to $y-x^3\ge0$, $y+x^3\le0$ and $y\ge0$.
The restrictions $y-x^3\ge0$, $y+x^3\le0$, and $y\ge0$ can be rewritten as $0\le y\le-x^3$ or, equivalently, as $x\le0\ \&\ 0\le y\le-x^3$. For each real $x\le0$, the minimum of $f(x,y)=(x-2)^2+y$ subject to $0\le y\le-x^3$ is $f(x,0)=(x-2)^2$. So, the minimum in question is $\min_{x\le0}(x-2)^2=(0-2)^2=4$.