# What is the slope of the tangent line of 1/y-x^3y-y= C , where C is an arbitrary constant, at (1,-2)?

$m = - \frac{24}{9}$
$- 1 {y}^{- 2} \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {x}^{2} y - {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
Substitute $x = 1$ and $y = - 2$. Then solve for $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{24}{9}$