$\Rightarrow F= - \frac{{dU}}{{dx}} = - 3k|x{|^2}$ ...(i)

Also, for $SHM $ $x = a\sin \omega \,t$ and $\frac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
$ \Rightarrow $ acceleration $ = \frac{{{d^2}x}}{{d{t^2}}} = - {\omega ^2}x$

$\Rightarrow F = ma$
$ = m\frac{{{d^2}x}}{{d{t^2}}} = - m{\omega ^2}x$ ...(ii)
From equation (i) & (ii) we get $\omega = \sqrt {\frac{{3kx}}{m}} $
$ \Rightarrow T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{{3kx}}} = 2\pi \sqrt {\frac{m}{{3k(a\sin \omega \,t)}}} $

$ \Rightarrow T \propto \frac{1}{{\sqrt a }}$.