Answer
Please see the work below.
Work Step by Step
As we know that
$E=\frac{1}{2}mv^2+\frac{1}{2}kx^2$
We differentiate both sides to obtain:
$\frac{dE}{dt}=\frac{1}{2}m\frac{d}{dt}v^2+\frac{1}{2}k\frac{d}{dt}x^2=0$
$\implies \frac{1}{2}m2v\frac{dv}{dt}+\frac{1}{2}2x\frac{dx}{dt}=0$
As $\frac{dx}{dt}=v$ and $\frac{dv}{dt}=a$
$\implies mva+kxv=0$
$v(ma+kx)=0$
$ma=-kx$
$\implies m\frac{d^2x}{dt}=-kx$
This is same as equation 13.3, hence proved.
