## Calculus (3rd Edition)

$\frac{dy}{dt}=(\frac{-x^{2}-y^{2}}{2xy})\frac{dx}{dt}$
Implicitly differentiating $x^{3}+3xy^{2}=1$, we have $3x^{2}\times\frac{dx}{dt}+3(\frac{dx}{dt}\times y^{2}+x\times2y\frac{dy}{dt})=0$ $\implies 3x^{2}\frac{dx}{dt}+3y^{2}\frac{dx}{dt}+6xy\frac{dy}{dt}=0$ $\implies 6xy\frac{dy}{dt}=-3x^{2}\frac{dx}{dt}-3y^{2}\frac{dx}{dt}$ $\implies 6xy\frac{dy}{dt}=(-3x^{2}-3y^{2})\frac{dx}{dt}$ $\implies \frac{dy}{dt}=(\frac{-3x^{2}-3y^{2}}{6xy})\frac{dx}{dt}$ $=(\frac{-x^{2}-y^{2}}{2xy})\frac{dx}{dt}$