take z-x=a x-y=b and so y-z=-(a+b)

substituting these values in the above eqn and cross multiplying u get

x^(b)=z^-(a+b).......eq(1)

x^(a)=y^-(a+b)......eq(2)

y^(b)=z^(a)

now substitute values of y^(b) and z^(a) in eq(1) and eq(2) we get

x^(b)=(yz)^(-b)

x^(a)=(yz)^(-a)

now solving this u get

x^(b-a)=(1/yz)^(b-a)

so log(x)=log(1/yz)... eq(3)

now coming to rhs take log on both sides u get

xlog(x)+ylog(y)+zlog(z)=0

using eq 3 and the fact that x=1/yz u get

log(yz)^(-yz)+log(yz)^(yz)=0

hence proved