Question

How do you implicitly differentiate `x^3 + y^3 = 9xy`?

Answer 1

I found: `(dy)/(dx)=(x^2-3y)/(3x-y^2)`

Explanation:

When deriving implicitly you must remember to trea `y` as a function of `x` so that, for example, if you have:
`y^2` derived gives you: `2y*(dy)/(dx)`
In your case you have:
`3x^2+3y^2(dy)/(dx)=9y+9x(dy)/(dx)` where on the right I used the Product Rule:

Collecting `(dy)/(dx)` you get:

`(dy)/(dx)=(3x^2-9y)/(9x-3y^2)=(cancel(3)(x^2-3y))/(cancel(3)(3x-y^2))=(x^2-3y)/(3x-y^2)`