#(xylogxy)/(x+y)=(yzlogyz)/(y+z)=(zxlogzx)/(z+x)#
#=>(logxy)/((x+y)/(xy))=(logyz)/((y+z)/(yz))=(logzx)/((z+x)/(zx))#
#=>(logx+logy)/(1/x+1/y)=(logy+logz)/(1/y+1/z)=(logz+logx)/(1/z+1/x)=k# say
then #k(1/x+1/y)=logx+logy# .........................(1)
#k(1/y+1/z)=logy+logz# .........................(2) and
#k(1/z+1/x)=logz+logx# .........................(3)
Adding the three, we get #2k(1/x+1/y+1/z)=2(logx+logy+logz)# or
#k(1/x+1/y+1/z)=logx+logy+logz# .........................(4)
Now subtracting (1), (2) and (3) from (4), we get
#k/z=logz# i.e. #k=zlogz=logz^z# .........................(5)
#k/x=logx# i.e. #k=xlogx=logx^x# .........................(6)
#k/y=logy# i.e. #k=ylogy=logy^y# .........................(5)
(5), (6) and (7) give us
#logx^x=logy^y=logz^z#
i.e. #x^x=y^y=z^z#
