We can calculate the first derivative separately for #x<0# and #x>0#:

1) For #x<0# we have:

#y=x-ln(-x)#

#(dy)/(dx) = 1-1/x#

2) For #x>0# we have:

#y=x-ln(x)#

#(dy)/(dx) = 1-1/x#

Thus the derivative is the same in the two intervals #(-oo,0)# and #(0,+oo)#. The second derivative is:

#(d^2y)/(dy^2) = d/(dx) (1-1/x)= 1/x^2#

We can therefore see that the function has only a critical point for #x=1# and that

#y'(x) < 0# for #x in (-oo,0) uu (0,1)#

#y'(x) >0# for #x in (1,+oo)#

so this critical point is a local minimum. It has no inflection points and is concave up in its domain.

We can also note that:

#lim_(x->-oo) y(x) = -oo#

#lim_(x->+oo) y(x) = +oo#

#lim_(x->0) y(x) = +oo#

So #y(x)# starts from #-oo# strictly increases approaching #+oo# for #x->0^-#, then decraeses starting from #+oo# as #x->0^+#, reaches a minimum for #x=1# and then starts increasing again approaching #+oo# for #x->+oo#.