calculate resonancefrequence RLC||C2
Because you've tried and seem honest I will give an answer.

There are two resonant frequencies. There is a the series resonant frequency which exclusively depends only on L and C1 and there is a parallel resonant frequency that depends on L and the combination of C1 and C2. Here's my take on it: -

The impedance is \$\dfrac{(R + sL + \frac{1}{sC_1})\frac{1}{sC_2}}{R + sL + \frac{1}{sC_1}+\frac{1}{sC_2}}\$

That's basically product over sum as you did.

It reduces to Z = \$\dfrac{s^2 + s\frac{R}{L} + \frac{1}{LC_1}}{sC_2(s^2 + s\frac{R}{L} + \frac{1}{L}(\frac{1}{C_1}+\frac{1}{C_2}))}\$

Half a sheet of regular notepad paper is all you need providing you aim for the standard format solution (yes I did use algebra rather than a math tool!).

So, the two resonant frequencies are \$\dfrac{1}{2\pi\sqrt{LC_1}}\$

And \$\dfrac{1}{2\pi\sqrt{LC_S}}\$ where \$C_S\$ is the two capacitors in series.

Those formulas numerically produce natural resonant frequencies of 6.059 kHz (series) and 12.831 kHz (parallel) and, if you look at my simulation below....



... it ties in with the math.

I made the above plot with R at 2 ohms just so that it would "amplify" the two resonant frequencies with minimal error because working with jw isn't quite the same as working with s when there is a fair bit of damping. The plot is a transfer function plot, not an impedance plot so don't be confused by the overall rising trend. The impedance will have an overall falling trend due to the "sC2" in the denominator.