Step-by-step Solution:

\[ f(x) = \lim_{{x}\rightarrow 0} \frac{{6}^x - {3}^x - {2}^x + 1}{\log_e 9 (1 - \cos x)}, \] we use Taylor series and L'Hôpital's Rule. For small \( x \), \( 1 - \cos x \approx \frac{x^2}{2} \), and \( a^x \approx 1 + x \ln a \). The numerator becomes: \[ 6^x - 3^x - 2^x + 1 \approx x (\ln 6 - \ln 3 - \ln 2) + \frac{x^2}{2} \left((\ln 6)^2 - (\ln 3)^2 - (\ln 2)^2\right). \] Since \( \ln 6 = \ln 2 + \ln 3 \), the first-order term vanishes, leaving: \[ \frac{x^2}{2} \left((\ln 6)^2 - (\ln 3)^2 - (\ln 2)^2\right). \] Substituting into the limit: \[ f(x) = \frac{(\ln 6)^2 - (\ln 3)^2 - (\ln 2)^2}{2 \log_e 3}. \] Simplify using \( (\ln 6)^2 = (\ln 2 + \ln 3)^2 \): \[ f(x) = \frac{2 \ln 2 \ln 3}{2 \log_e 3} = \ln 2. \]