Answer
$\frac{dy}{dx} = \frac{-3t^2y-2y^3}{(6y^2t + t^3)cos(t)}$
Work Step by Step
By the chain rule, $\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx}$. Thus, because we are given $\frac{dt}{dx} = \frac{1}{cos(t)}$, we only need to find $\frac{dy}{dt}$. We can accomplish this by applying implicit differentiation and solving for $\frac{dy}{dt}$:
$2y^3t+t^3y = 1$
$6y^2t *\frac{dy}{dt} + 2y^3 + 3t^2y + t^3 * \frac{dy}{dt} = 0$
$(6y^2t + t^3)\frac{dy}{dt} = -3t^2y-2y^3$
$\frac{dy}{dt} = \frac{-3t^2y-2y^3}{6y^2t + t^3}$
Thus, $\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx} = \frac{-3t^2y-2y^3}{6y^2t + t^3} * \frac{1}{cos(t)} = \frac{-3t^2y-2y^3}{(6y^2t + t^3)cos(t)}$
