Using differential equations to find an implicit curve

This differential equation is simply the Product Rule of:

$\frac{d}{dx}(y^3 \cdot sin(x)) =0 \Rightarrow \boxed{y^3 \cdot sin(x) = C}.$

When doing implicit differentiation I was considering how if you had a derivative in the form $\frac{dy}{dx}$ =f(x,y) how you could find the implicit curve like how you could use the fundamental theorem of calculus to find an explicit curve from a derivative. This is why I have posted a relatively easy differential equation to solve.

$\frac{dy}{dx}=-\frac{y^3cosx}{3y^2sinx}$ $-\frac{3}{y}dy= \cot x dx$ . Integrating both sides $lny^{-3}=\ln \sin x +c$ $y^{-3}=e^c \sin x$ $\Rightarrow p=y^3 \sin x$