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