Answer
The stability location of the electron is at x = 0 & this corresponds to an unstable equilibrium.
Work Step by Step
$U(x)=ax^{2}-bx^{4}-(1)$
Equilibria occur where the potential energy has a maximum or minimum - that is, where its derivative is zero. Taking the derivative of U(x) & setting it to zero gives,
$0=\frac{dU}{dx}= 2ax-4bx^{3}=2x(a-bx^{2})$
$x=0$
$\space\space\space or$
$a-2bx^{2}=0$
$x=\pm \sqrt {\frac{a}{2b}}$ ; Let's plug known values into this equation.
$x=\pm\sqrt {\frac{-8\space aj/nm^{2}}{1\space aj/nm^{4}}}= \pm \sqrt {-8}\space nm$
This is not a real number. So this solution should neglect.
Therefore the location of the stability is x = 0
Let's find the value of $\frac{d^{2}U}{dx^{2}},$ to find the state of equlibrium.
$\frac{d^{2}U}{dx^{2}}=2a-12bx^{2}=-16\space aj/nm^{2}-12x^{2}\space aj/nm^{4}$
This value is negative for all x values ; $\frac{d^{2}U}{dx^{2}}\lt0$
So this corresponds to an unstable equilibrium