Question: find dy/dx using implicit differentiation: y= 3xy - 2x^3.
From y = 3xy - 2x^3
y(1 - 3x) = -2x^3
y = -2x^3/(1-3x)
Writing y' for dy/dx,
Differentiate (implicitly) y = 3xy - 2x^3
y' = 3(xy' + y) - 6x^2
y' = 3xy' + 3y - 6x^2
y'(1 - 3x) = 3(-2x^3/(1-3x)) - 6x^2
y' = -6x^3/(1-3x)^2 - 6x^2(1-3x)/(1-3x)^2 = {-6x^3 - 6x^2 + 18x^3)/(1-3x)^2
y' = 6x^2(2x - 1)/(1-3x)^2