Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 51

The vertical asymptotes of the given function are x = 0 and x = e, such that:
lim_(x->0) (ln(1 - ln x)) = ln lim_(x->0) (1 - ln x) = ln (1 - lim_(x->0) ln x)
lim_(x->0) (ln(1 - ln x)) = ln (1 - (-oo))
lim_(x->0) (ln(1 - ln x)) = ln oo = oo
Hence, the function has vertical asymptote x = 0.
lim_(x->e) (ln(1 - ln x)) = ln (1 - lim_(x->e) ln x)
lim_(x->e) (ln(1 - ln x)) = ln (1 - ln e)
lim_(x->e) (ln(1 - ln x)) = ln (1 - 1)
lim_(x->e) (ln(1 - ln x)) = -oo
Hence, the function has vertical asymptote x = e.
You need to evaluate the horizontal asymptotes of the function:
lim_(x->oo) (ln(1 - ln x)) impossible to be evaluated since 1 - ln x < 0 as x approaches to oo.
b) You need to evaluate the monotony of the function, hence, you need to determine the intervals for f'(x)>0 or f'(x)<0.
You need to determine the derivative of the function:
f'(x) = (ln(1 - ln x))' => f'(x) = (1/(1-ln x))*(1 - ln x)'
f'(x) = (-1/x)/(1-ln x)
You need to notice that f'(x) > 0, hence the function increases, for x in (e,oo) and f'(x) < 0, hence the function decreases, for x in (0,e).