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 [itex]H(j \omega)[/itex]. 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 [itex]H(j \omega)[/itex] (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 [itex]H(j \omega)[/itex] (assuming you got right the first expression). Defining [itex]R_{tot} = R+R_L[/itex], you should get:
[tex]H(j \omega) = \frac{j\omega RC}{1+j\omega R_{tot} C -\omega^2 LC}[/tex]
Now, it is customary to simplify this expression further by explicitly writing it in Bode form. In other words, you write it as:
[tex]H(j \omega)=\frac{j2\zeta' \frac{\omega}{\omega_0}}{1+j2\zeta \frac{\omega}{\omega_0} - \frac{\omega^2}{\omega_0^2}}[/tex]
where [itex]\zeta' , \zeta , \omega_0[/itex] are all known. At this point, we are half-way. Now, we must consider: [itex]|H(j \omega)|[/itex] (question: can we, instead, consider [itex]|H(j \omega)|^2[/itex] to simplify things? If so, why?). To calculate it, I suggest you to use: [itex]x=\frac{\omega^2}{\omega_0^2}[/itex] and do all the math with respect to [itex]x[/itex]. Stated this way, you should be able to conclude it.