Question

How do you differentiate `-y=lnxy-xy`?

Answer 1

`y' = - (y - xy^2)/( xy + x - x^2y )`

Explanation:

this is a bit easier [administratively] if you use the Implicit Function Theorem...

which says that `color{red}{y' = - (f_x)/(f_y)}` for function `f(x,y) = const`

so from
`-y=lnxy-xy`

we get
`y + lnxy-xy \color{blue}{= 0}= f(x,y)`

and so....
`f_x = 1/(xy)*y - y`
`= 1/x - y`
and
`f_y = 1 + 1/(xy)*x - x`
`= 1 + 1/(y) - x`

so using the IFT....
`y' = - (f_x)/(f_y)`

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

`= - (y - xy^2)/( xy + x - x^2y )`

there are some pretty simple ways to intuit the Implicit Function Theorem

one is to note that if `f(x,y) = c`

then the total differential `df = 0`

and `df = f_x dx + f_y dy = 0`

so `dy/dx = - (f_x)/(f_y)`