We have the function $$f_a(x)=\frac{a-x}{\ln (a-x)}, \ a\in \mathbb{R}$$
To get the domain of that function we have to consider the following restrictions: $$\begin{cases}\ln (a-x)\neq 0 \\ a-x>0 \end{cases} \Rightarrow \begin{cases} a-x\neq 1 \\ a-x>0 \end{cases} \Rightarrow \begin{cases} x\neq a-1 \\ x<a \end{cases}$$ So the domain is $$D_f=\{x\in \mathbb{R}: x<a \ \text{ and } x\neq a-1\}$$ Is this correct?
The function has to root since $f_a=0 \Rightarrow a-x=0 \Rightarrow x=a\notin D_f$, right?
Next, I want to determine the extremas. The first derivative is $$f_a'=\frac{-\ln (a-x)+1}{\left (\ln (a-x)\right )^2}$$ The root of the first derivative is $x=a-e$.
To check if this is a minimum or a maximum we need the second derivative, which is at $x=a-e$ positive, so we have a minimum.
Is everything correct so far?
Is this the only extremum or do we have to check also the points where the functions is not defined, i.e. $x=a$ and $x=a-1$ ?
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$, $$\lim_{x\to -\infty}f(x),\quad \lim_{x\to (a-1)^-}f(x),\quad \lim_{x\to (a-1)^+}f(x),\quad \lim_{x\to a^-}f(x).$$ Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.