I wish to find $y$ in terms of $x$ (not necessarily exact, just upto some order, eg. $y=cx^2$, where $c$ is a constant)
$x$ and $y$ follow this relation -
$$\frac{x^2}{2y} - x + y - y\Big(1-\frac 1{y}\Big)^x \leq \frac 1{2}$$
(The RHS of inequation can be any constant between $0$ and $1$, I chose $\frac 1{2}$ for simplicity)
Any leads as to how to simplify this is appreciated.
Note by Bernoulli’s inequality, $\displaystyle \left(1-\frac1y\right)^x>1-\frac{x}y$ and the inequality gets better with large $y>x$, so the question reduces to $\dfrac{x^2}{2y}\leqslant \dfrac12 \iff x^2\leqslant y$ asymptotically.
This is a region, bounded on the right (and below) by $y=x^2$.