What is the derivative y' of x + y + ln(xy) = 0

Differentiate x + y + ln(xy) =
0

`dx + dy + 1/(xy) d(xy) = 0` Using the chain
rule

d(xy) = xdy+ydx using the product
rule

`dx+dy+1/(xy)(xdy+ydx)=0`

Distributing
`1/(xy)` we get

`dx + dy + dy/y + dx/x =
0`

Moving the dx terms to the right
side

`dy + 1/y dy = - (dx + 1/x
dx)`

Factoring dx and dy we
get

`dy(1+1/y) =
-(1+1/x)dx`

`dy/dx = -(1+1/x)/(1+1/y) =
-((x+1)/x)/((y+1)/y)=-(y(x+1))/(x(y+1)) `