y=ce^(y/x) , where c is a constant find y' ?

Answer 1

y = ce^(y/x)
ln y = ln (ce^(y/x))
ln y = ln c + ln e^(y/x)
ln y = ln c + (y/x)
Differentiate both side d/dx
(1/y)(dy/dx) = 0 + (dy/dx)(1/x) - y/x^2
(dy/dx)(1/y - 1/x) = -y/x^2
(dy/dx)((x-y)/xy) = -y/x^2
dy/dx = y' = (-y^2)/(x(x-y))

Answer 2

y= c.e^(y/x)
dy/dx = {c.[x(dy/dx) - y].e^(y/x)} / (x^2)

use the chain rule and evaluate the derivative of (y/x) using the quotient rule.
Hope this helps!