An inductor is in series with the parallel connection of inductor - capacitor. How to find the resonance Freq?

F=2piLxl

That is frequency is equel to two times pi times the inductance times the inducive reactance of the circuit.

You can estimate this  by looking at it as a parallel tuned circuit and a series tuned circuit.
The C and L in parallel will resonate at a frequency given by:
F^2 = 25330 / (L * C) where F is MHz , L is uH and C is pF

This will be a maximum impedance point.

The single L will also resonate in series with the C at a frequency given by the same equation
F^2 = 25330 / (L * C) where F is MHz , L is in uH and C is in pF

This will be a minimum impedance point.

Don't forget to take the square root to get frequency.

eg a 20uH inductance is placed in series with a 30 uH coil which has a 1000 pF capacitor across it.

Series circuit :
F = SQRT( 25330/(20 * 1000)) =1.125 MHz

Parallel Circuit:
F = SQRT ( 25330 / (30 * 1000)) = 0.919 MHz

series tunned=s/c during resonant

so find the frequency that will gave j2pfL+(1/j2pfL-j2pfC)^-1=0

Assume L1 is connected in series with the parallel combination of L2 and C.

 I can not think of a use for a circuit like this one but here is how I would attempt to find the resonant frequency. Please do not bet your last dollar that this is the correct or easiest way to solve for resonance.

The total reactance (Xt) of the parallel combination of C and L2 = 1/ [(B of C) + (B of L2)] where (B) symbolizes the susceptance of L2 and C.  Susceptance is the reciprocal of reactance (X) and susceptance values can be added together to get the total suseptance of a capacitor and inductor connected in parallel. Then you can take the reciprocal of the total susceptance to get back to total reactance.
Here are the reactance formulas:  X of C = 1 / 2pi fC and X of L = 2pi fL. Of course the susceptance formulas (B) of C and (B) of L are the reciprocals of these formulas.

The frequency at which  (X of L1) = (Xt) is the resonant frequency.  Plug in the values for (L1), (L2) and (C) on each side of the equation and solve for the resonant frequency (f).