Show that $\displaystyle xy^3 - xy^3\sin{(x)} = 1$ is an implicit solution to
$\displaystyle \frac{dy}{dx} = \frac{(x\cos{(x)} + \sin{(x)} - 1)y}{3(x - x\sin{(x)})}$
on the interval $\displaystyle \left(0,\frac{\pi}{2}\right)$
Differentiate both sides:
$\displaystyle \frac{d}{dx} (xy^3 - xy^3\sin{(x)}) = \frac{d}{dx}1$
$\displaystyle \frac{d}{dx} (xy^3) - \frac{d}{dx}( xy^3\sin{(x)}) =0$
Product rule on the 2nd term!
$\displaystyle \frac{d}{dx} (xy^3) - sin(x) \frac{d}{dx}( xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0$
Take out a factor of $\displaystyle \frac{d}{dx} (xy^3) $ on LHS:
$\displaystyle (1-\sin{(x)}) \frac{d}{dx} (xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0$
Using implicit differentiation with the product rule: $\displaystyle \frac{d}{dx} (xy^3) = y^3 + 3y^2 x \frac{dy}{dx} $, and also $\displaystyle \frac{d}{dx} \sin{(x)} = cos(x)$ hence:
$\displaystyle (1-\sin{(x)})( y^3 + 3y^2 x \frac{dy}{dx}) - xy^3 \cos{(x)}) =0$
Now rearrange to get into explicit form, and then manipulate to get it into the form posed by the question!