Find (dy/dx) of x^3 - x + y^3 = 6 + 2y^2 in terms of x and y

In order to differentiate this expression, we need to use implicit differentiation. An expression in the form y = f(x), where f(x) means "a function of x", is called an explicit equation and needs to be differentiated explicitly (i.e. the explicit equation y= x² differentiates to 2x). However, in the above question, the equation consists of functions of both y and x (the functions of y are y³ and 2y² and those of x are x³ and -x) meaning that it is an implicit equation.
Firstly we need to differentiate the functions of x: x³ differentiates to 3x² and -x differentiates to -1. We then need to differentiate the functions of y: y³ differentiates to 3y² and 2y² differentiates to 4y. However, as we need to differentiate y in terms of x (given by dy/dx in the question), the correct differentiated terms are 3y²(dy/dx) and 4y(dy/dx). Finally, the constant 6 differentiates to 0 as it has no x or y terms. Replace the terms in the question with the differentiated terms to get: 3x² - 1 + 3y²(dy/dx) = 0 + 4y(dy/dx). Now we need to rearrange the equation to get all terms with (dy/dx) on one side. This leaves us with: 3y²(dy/dx) - 4y(dy/dx) = 1 - 3x² . If we factorise (dy/dx) out of the left hand side, we get: (dy/dx)(3y²- 4y) = 1 - 3x². Finally, to get dy/dx in terms of x and y, we need to divide both sides of the equation by (3y²- 4y). This gets us to the answer of the question which is (dy/dx) = (1 - 3x²)/(3y²- 4y).