If the function \( f(x) = \frac{e^{x}\left(e^{\tan x - x}-1\right)+\log_{9}(\sec x + \tan x)-x}{\tan x - x} \) is continuous at \( x = 0 \), then the value of \( f(0) \) is equal to:
\( 2 \)
\( \frac{2}{3} \)
\( \frac{1}{2} \)
\( \frac{3}{2} \)