A particle free to move along the x-axis has potential energy given by
$U\left(x\right)=k\left[1-\exp \left(-x^2\right)\right]$ for $-\infty \le x\le +\infty$, where k is a constant of appropriate dimensions. Then

Figure shows the plot of U(x) versus x.
At x=0, potential energy $U\left(0\right)=k[1-\exp (0\left)\right]=k(1-1)=0$ and it has a maximum value =k at $x=\pm \infty$ since
$U(\pm \infty )=k\left[1-\exp (-\pm \infty {)}^2\right]=k(1-0)=k$
Since the total mechanical energy has a constant value =(k/2), the kinetic energy will be maximum at x=0 and minimum at $x=\pm \infty$. At x=0
${\left(\frac{dU}{dx}\right)}_{x=0}={\left[2kx\exp \left(-x^2\right)\right]}_{\mathrm{at}x=0}=0$

Hence the particle is in stable equilibrium at x=0 (origin) and would oscillate about x=0 (for small displacements) simple harmonically.