Question

How do you find the derivative of y in the equation `ln(xy)=x+y`?

Answer 1

You have to remember that in this "implicit" function `y` is a function of `x` so deriving you get:
`1/(xy)*(y+xdy/dx)=1+dy/dx` where you used the Chain Rule on the `ln`
Then:
`1/x+1/ydy/dx=1+dy/dx`
So you collect `dy/dx` and get:
`dy/dx=(1-1/x)/(1/y-1)=((x-1)/x)(y/(1-y))=(y(x-1))/(x(1-y)`

Hope it helps