Cannot figure out resonance of an LRC circuit
we are looking for the natural frequency, and two corner frequencies
  (where Vo/Vin = 0.707)


The natural resonant frequency is purely determined by L and C and is when the impedance of L equals the impedance of C. Why not write those two formula down and equate then solve for frequency. This part is quite simple.

The two corner frequencies I take as meaning the -3dB points of the band-pass filter - are you aware that what you have described IS a BPF? Anyway this is more onerous mathematical solution but you should end up with a formula that generally is: -

TF = \$\dfrac{2\zeta\omega_n s}{s^2 + 2\zeta\omega_n s + \omega_n^2}\$

I use the term "general" because any 2nd order filter be it mechanical or electronic uses the same general terms although for an electronic circuit the terms R, L and C will be part of the formula.

Where s = jw and \$\zeta\$ = 1/2Q (do you understand Q?). 

\$\omega_n\$ is the natural resonant frequency in radians per second.

Q (or inversely \$\zeta\$) dictate how "fat" the response is in the frequency domain. For instance a very high Q gives a very sharp response with the -3dB points very close in frequency. Another definition of Q is that Q = natural resonant frequency divided by -3dB bandwidth.

Hope this helps.