What have I done wrong in determining the transfer function for this RCL circuit
Jan's answer gives you the result, but the reason you didn't get the correct answer is because you didn't consider the intermediary stage.



simulate this circuit &ndash; Schematic created using CircuitLab
Given your approach, the first thing to do is determine the voltage at point a by considering the whole L || (R2 + C) as an impedance, and only then use that voltage with the divider formed by R2 and C. I won't expand the equations, it looks like you're good with them:
$$\begin{align}
Z_{RC}&amp;=R_2+\dfrac{1}{sC}\tag{1} \\
Z_a&amp;=\dfrac{1}{\dfrac{1}{sL}+\dfrac{1}{Z_{RC}}}\tag{2} \\
V_a&amp;=V_1\cdot\dfrac{Z_a}{Z_a+R_1}\tag{3} \\
V_x&amp;=V_a\cdot\dfrac{\dfrac{1}{sC}}{\dfrac{1}{sC}+R_2}\tag{4}
\end{align}$$
And you get the correct transfer function. When you'll be expanding the above, you'll realize that it can get messy, that's why for these cases, it's better to use KCL/KVL/FACTS/etc, because the resulting system of equations is easier to analyze. In the end, it looks like you could have done this yourself if you payed a bit more attention.

Given your comment, I'll try to expand a bit. You have two things: the way you want to solve this is by using the voltage dividers, and the circuit is a Cauer network (also called ladder). Since you're interested in finding out the voltage at the output (V(x)), you first need to find out the previous node voltage, V(a). For that, you have to look at the circuit from a different perspective:



simulate this circuit
V(a) divides the input voltage and it does so at point a, so that's  how the circuit looks like from the input's perspective (blue arrow). Which means to calculate V(a) you first need to determine the equivalent impedance formed by L, R2, and C, which is R2 in series with C (1), all in parallel with L (2). Expanded and continued from the equations above:
$$\begin{align}
Z_a&amp;=\dfrac{1}{\dfrac{1}{sL}+\dfrac{sC}{sR_2C+1}}=\dfrac{s^2R_2LC+sL}{s^2LC+sR_2C+1}\tag{5} \\
V_a&amp;=V_1\cdot\dfrac{s^2R_2LC+sL}{s^2(R_1+R_2)LC+s(R_1R_2C+L)+R_1}\tag{6}
\end{align}$$
And it's this (6) that needs to be used in (4) in order to find V(x), so you can see, it gets pretty messy. If you compare these with KCL/KVL/FACTS (in Verbal Kint's answer), you can't really say that they are less messy. BTW, Verbal Kint used a different approach by using the Thevenin equivalent source and then the output (also hinted at by Chu's comment), which means he went from output towards the input; different way, same results. Anyway, if you plot this transfer function agains V(a), you will get the same response:

I have kept the default Rser=1m for the inductances because it makes the comparison of the results better (otherwise they would have completely overlapped).