## Calculus 8th Edition

For $x^3-xy^2+y^3=1$ find $\frac{dy}{dx}$
Differentiate both side with respect to x $3x^2-y^2-2xy\frac{dy}{dx}+3y^2\frac{dy}{dx} =0$ Isolate $\frac{dy}{dx}$ $\frac{dy}{dx}(3y^2-2xy)=y^2-3x^2$ $\frac{dy}{dx}=\frac{y^2-3x^2}{(3y^2-2xy)}$