# How do you differentiate y^3-y=x^3-x?

$y ' = \frac{3 {x}^{2} - 1}{3 {y}^{2} - 1}$
$\left({y}^{3} - y = {x}^{3} - x\right) '$
$3 {y}^{2} y ' - y ' = 3 {x}^{2} - 1$
$y ' \left(3 {y}^{2} - 1\right) = 3 {x}^{2} - 1$
$y ' = \frac{3 {x}^{2} - 1}{3 {y}^{2} - 1}$