Sum the two terms:
#1/x-1/(e^x-1) = (x-e^x+1)/(x(e^x-1))#
The limit is now in the indeterminate form #0/0# so we can now apply l'Hospital's rule:
#lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) (d/dx (e^x+1-x))/(d/dx x(e^x-1))#
#lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) (e^x-1)/(e^x-1+ xe^x)#
and as this is till in the form #0/0# a second time:
#lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) (d/dx (e^x-1))/(d/dx (e^x-1+ xe^x))#
#lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) e^x/(e^x +xe^x+ e^x)#
#lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) 1/(x+2) = 1/2#
graph{1/x-1/(e^x-1) [-10, 10, -5, 5]}