# How do you differentiate (y^2-1)^2+x=0?

$y ' = \frac{1}{4 y \left(1 - {y}^{2}\right)}$
$x = - {y}^{4} + 2 {y}^{2} - 1$
$\frac{\mathrm{dx}}{\mathrm{dy}} = - 4 {y}^{3} + 4 y = 4 y \left(1 - {y}^{2}\right)$.
$y ' = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \frac{1}{4 y \left(1 - {y}^{2}\right)}$