How do you find the derivative of ln(xy)=x+y

implicit differentiation

1/(xy) (xy)' = 1 + y'
1/(xy) (y + xy') = 1 + y'
y + xy' = xy + xyy'
y'(x-xy) = xy-y
y' = y(x-1) / x(1-y)

or,

ln(xy) = x+y
lnx + lny = x+y
1/x + 1/y y' = 1 + y'
y'(1/y - 1) = 1 - 1/x
y' = (1 - 1/x) / (1/y - 1)
as above

or,
ln(xy) = x+y
xy = e^(x+y) = e^x * e^y
y + xy' = e^x + e^y y'
y' = (e^x-y)/(x-e^y)