Question

How do you implicitly differentiate `y-(3y-x)^2-1/y^2=x^3 + y^3- xy`?

Answer 1

By differentiating `x` and `y` like your usual differetiation practices.

Explanation:

`y-(3y-x)^2-1/y^2=x^3+y^3-xy`

I will let `(dy)/(dx)=y'`

Now begin the implicit differentiation

`y'-2(3y-x)(3y'-1)-(-2)1/y^3y'=3x^2+3y^2y'-(y+xy')`

I moved all the terms to one side and factored out `y'`

`y'[7-18y+2/y^2-3y^2+x]+7y-2x-3x^2=0`

Trick is; after differentiating any form of `y`, add `y'`

Example: `d/dxy^2=2yy'`

Hope this helps. Cheers