1     1      1  
    xy + ——— + ——— = ———— + x + y.
          x     y     xy

xy+(1/x)+(1/y)=(1/xy)+x+y
x^2y^2+x+y=1+x^2y+xy^2
x^2y^2-1=xy(x+y)-(x+y)
(xy+1)(xy-1)=(xy-1)(x+y)
thus we have 2 groups of solutions either
xy=1 thus y=1/x
if xy!=1 then we have
xy+1=x+y
(x-1)y=(x-1)
thus either x=1 or y=1

so in summary all solutions are of the form
(1,y) (x,1) or (x,1/x)
with obvious restriction that neither x or y are zero