Tricky problem! Do it!

first term:
(1+x+1/y) = (xyz+xyz*x+xz) = xz(y+xy+1) = xz(1+y+xy) second term: (1+y+1/z) = (1+y+xy) third term: (1+z+1/x) = (xyz+z+yz) = z(xy+1+y) = z(1+y+xy)

we know xyz=1 so, 1/xz=y and 1/z=xy

now making inverse then adding we get, 1/{xz (1+y+xy)} + 1/(1+y+xy) + 1/{z (1+y+xy)} = [ y/(1+y+xy) + 1/(1+y+xy) + xy/(1+y+xy)]
= [ (y+1+xy)/(1+y+xy) ] =1

if we remove z by 1/xy for all z then the ans will b 1

Tried to nicely format the answer so others can follow it : $(1+x+y^{-1})^-1 + (1+y+z^-1)^-1 + (1+z+x^-1)^-1$

For ease of reading reform this as temporarily $a^-1 + b^-1 + c^-1$

with a only (and using xyz =1) $1 +x+y^-1 = xyz + xyz.x + zx = xz.(y+xy1)$

with b only (and using xyz =1) $1 +y+z^-1 = 1 + y + xy$

with c only (and using xyz =1) $1 +z+x^-1 = xyz + z + yz = z(xy+1+y)$

Recombining : $xz.(1+y+x.y)^-1 + (1+y+x.y)^-1 + z.(1+y+x.y)^-1 =$ $y/(1+y+x.y) + 1/(1+y+x.y) + x.y/(1+y+x.y) =$ $(1+y+y.x)/(1+y+y.x) =$ $\boxed{1}$

it can be done this way too... (1+x+y-1)-1 lets replace y-1 by xz(since xyz=1 implies xz=y-1) Now we have (1+x+xz)-1, in the last term we have the term with same x and z variables so let’s combine both. Before that lets modify that as well(1+z+x-1)-1 implies x/(1+xz+x) also (1+x+xz)-1 is 1/(1+x+xz) so this becomes (1/(1+x+xz)) + (x/(1+x+xz)) = (1+x)/(1+x+xz) now the term remaining(ie (1+y+z-1)-1 has y and z terms, lets converts both of them interms of x and y(it can be any combinations depending on our own convenience) so (1+y+z-1)-1 in terms of x and y implies (1+y+xy)-1 = 1/(1+y+xy) and (1+x)/(1+x+xz) in terms of x and y implies (1+x)/(1+x+y-1) = y(1+x)/(1+y+xy) =(y+xy)/(1+y+xy) At last (1/(1+y+xy)) + (y+xy)/(1+y+xy) = (1+y+xy)/(1+y+xy) = 1

Substitue xy in place of z, then the ans will be 1.