I'm afraid to say that the title is a bit misleading, since your actual problem is finding the resonance frequency of an ideal circuit for which you have the expression of H(j \omega). The fact that you started with a non-ideal inductor, a non-ideal capacitor or "non-ideal anything" (for that matter) doesn't anymore have a role to play in your equations.
I didn't precisely understand your schematic (can you find a picture of it?), thus I will have to assume that you got the right expression for H(j \omega) (but see below). If so, then you are dealing with a variation of the standard RLC circuit. I strongly suggest you to go back to your textbooks and see how the resonance frequency for RLC was calculated - it is almost exactly the same here! The calculations aren't actually harder, if you carefully deal with them. However, it seems to me you made some mistake in writing H(j \omega) (assuming you got right the first expression). Defining R_{tot} = R+R_L, you should get:
H(j \omega) = \frac{j\omega RC}{1+j\omega R_{tot} C -\omega^2 LC}
Now, it is customary to simplify this expression further by explicitly writing it in Bode form. In other words, you write it as:
H(j \omega)=\frac{j2\zeta' \frac{\omega}{\omega_0}}{1+j2\zeta \frac{\omega}{\omega_0} - \frac{\omega^2}{\omega_0^2}}
where \zeta' , \zeta , \omega_0 are all known. At this point, we are half-way. Now, we must consider: |H(j \omega)| (question: can we, instead, consider |H(j \omega)|^2 to simplify things? If so, why?). To calculate it, I suggest you to use: x=\frac{\omega^2}{\omega_0^2} and do all the math with respect to x. Stated this way, you should be able to conclude it.