If yx = ey-x, prove that dydx = (1+logy)2/logy

yx = ey-x
Taking logarithm on both sides,
x logy = y - x
Differentiate with respect to x,
⇒ logy + x.(1/y)dy/dx = dy/dx
From equation (1),
xy = 1/(1+logy)
⇒ logy + 1 = dy/dx(1 - xy)
⇒ logy + 1 = dy/dx(1 - 1/(1+logy))
⇒ logy + 1 = dy/dx(logy/(1+logy))
⇒ dy/dx = (1+logy)2/logy
Hence proved.