How can this Parallel LC circuit act as a filter?
It looks like a resonance circuit. It may be simpler to ignore the input voltage. 

This filter network will have to be in terms of current because they share the same voltage nodes. 

The output voltage will be the current divided by the conductance. 

\$V_{out} = \frac{i}{Y}\ = \frac{i}{G + j\omega C + \frac{1}{j\omega L}} \$

\$ V_{out} = \frac{i}{G + j\omega C(1 - \frac{1}{\omega^2 LC})} \$

When  \$ 1 - \frac{1}{\omega^2 LC} = 0 \rightarrow \omega^2 = \frac{1}{LC}\$ 

This gives a resonance curve looking something like 

https://commons.wikimedia.org/wiki/File:Universal_Resonance_Curve.svg

with the peak at when \$ \omega^2 = \frac{1}{LC}\ \$which is a value of \$ \frac{i}{G} \$ at the resonant frequency.

The bandwidth is measured between the half power points. These are the two 3dB lower than the peak. RLC circuits have a very narrow bandwidth. 

The -3dB amplitude will be \$10^{\frac{(20log(\frac{i}{G}) - 3)}{20}}\$ which will eventually too \$ \frac{i}{G} \frac{1}{10^{\frac{3}{20}}}  \$ which is very close too \$ \frac{i}{G} \frac{1}{\sqrt2}  \$.

If you follow the same logic, but with a series RLC circuit, then you will come across a nearly identical derivation. In this case we use impedance instead of admittance and are measuring the output voltage across \$R_L \$ for a corresponding input voltage.