If xy = ex-y, prove that dy/dx = logx/(1+logx)2

Given: xy = ex-y
Taking log on both the sides,
log(xy) = log(ex-y)
ylog(x) = (x-y)log(e) = (x-y) [log(e) = 1]
y = x/(1+log(x))
Differentiating w.r.t. x both the sides,
dy/dx = d/dx(x/(1+log(x)))
dy/dx = ((1+logx)d(x)/dx - x.d(1+log(x))/dx)/(1+log(x))2
dy/dx = ((1+logx)(1) - x (0 + 1/x))/(1+log(x))2
dy/dx = log(x)/(1+log(x))2
Hence proved.