Functions

$\begin{aligned} \chi & = f \left(x^2 \right) f \left(y^2 \right) - \frac 12 \left( f \left(\frac {x^2}{y^2} \right) + f \left(x^2 y^2\right)\right) \\ & = \cos \left(\log x^2 \right) \cos \left(\log y^2 \right) - \frac 12 \left( \cos \left(\frac {\log x^2}{\log y^2} \right) + \cos \left(\log \left(x^2 y^2\right) \right)\right) \\ & = \cos \left({\color{#3D99F6}2 \log x} \right) \cos \left({\color{#D61F06}2 \log y} \right) - \frac 12 \left( \cos \left({\color{#3D99F6}2 \log x} - {\color{#D61F06}2 \log y} \right) + \cos \left({\color{#3D99F6}2 \log x} + {\color{#D61F06}2 \log y} \right)\right) \\ & = \cos {\color{#3D99F6}\alpha} \cos {\color{#D61F06}\beta} - \frac 12 \left( \cos \left({\color{#3D99F6}\alpha} - {\color{#D61F06}\beta} \right) + \cos \left({\color{#3D99F6}\alpha} + {\color{#D61F06}\beta} \right)\right) \\ & = \cos \alpha \cos \beta - \frac 12 \left( \cos \alpha \cos \beta + \sin \alpha \sin \beta + \cos \alpha \cos \beta - \sin \alpha \sin \beta \right) \\ & = \cos \alpha \cos \beta - \cos \alpha \cos \beta \\ & = 0 \end{aligned}$

Therefore, the answer is $\boxed{\text{None of the others}}$ .