Is it possible to derive the angular frequency of a simple harmonic oscillator using total energy? 
I want to show that 
  $$\omega=\sqrt{\frac{k}{m}}$$
  using the fact that
  $$E=K+U=\frac{1}{2}mv_x^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2.$$

The issue is that I have derived a formula that isn't correct:
I first took the time derivative of the energy equation,
$\dfrac{dE}{dt}=mv_x\dfrac{dv_x}{dt}+kx\dfrac{dx}{dt}=kA\dfrac{dA}{dt}$
$\dfrac{dE}{dt}=mv_xa_x+kxv_x=kAv_x$.
So we now have 
$mv_xa_x+kxv_x=kAv_x$
$ma_x+kx=kA$
Substituting $a_x=-\omega^2x$ gives
$-m\omega^2x+kx=kA$
$m\omega^2x=k(x-A)$
$\omega^2=\dfrac{k}{m}\dfrac{(x-A)}{x}$
$\omega=\sqrt{\dfrac{k}{m}\dfrac{(x-A)}{x}}$.
This is very nearly correct but not quite...
Is it possible to prove this formula in this way? Also, if I haven't done something that's completely incorrect, then is this a standard formulae that I just haven't seen before?
 A: Energy is conserved in a simple harmonic oscillator, so $dE/dt = 0$.  From there you can get the angular frequency by your same logic.
Another way of seeing this: A is a constant of the motion, so its time derivative is 0.
A: Whenever you want to prove something, you need to get your hypotheses straight. I understand that you want to prove that a mass $m$ on a spring with constant $k$ moves harmonically with frequency $\sqrt{k/m}$, but you seem to be assuming that result in your "proof":
First off, writing $E = \frac12 k A^2$ already implies that the motion is harmonic with frequency $\sqrt{k/m}$, partly because you need to know $x(t)$ and $\dot{x}(t)$ to plug into $E$, but also because $A$ is the amplitude, a notion which doesn't make sense if the motion isn't harmonic!
Also, $dA/dt$ isn't $v$ (because $dx/dt$ is $v$), it's zero: the amplitude is a constant. Using that you can get the equation of motion, but there's that same problem again: stating that $\ddot{x} = -\omega^2 x$ already assumes that $x$ is a simple harmonic motion.
The way to prove what you want is to guess a solution of the form $x=\cos(\omega t)$ (you could also use $\sin(\omega t)$) and plug it either into the energy (and then demand that the energy be a constant, that is, $dE/dt = 0$), or into the equation of motion $\ddot{x} + \frac{k}{m}x =0$. You will find that for the equations to be satisfied, it must be that $\omega^2 = \frac{k}{m}$.
