Solve $3xdy = y(1+x\sin(x) - 3y^3 \sin(x))dx$ Solve $3xdy = y(1+x\sin(x) - 3y^3 \sin(x))dx$
Mt Attempt
$$3xy' = y(1+x\sin(x) - 3y^3 \sin(x)) \rightarrow 3xy'-y(1+x\sin(x)) = -3y^3\sin(x)$$
$$y'-y\frac{(1+x\sin(x))}{3x}= -\frac{y^3\sin(x)}{x}$$
To me, this is the Bernoulli form. Hence let $u = y^{1-3} = y^{-2}$ and $u' = -2y^{-3}y'$
Divide by $y^3$
$$\frac{1}{y^3}y'-\frac{1}{y^2}\frac{(1+x\sin(x))}{3x}= -\frac{\sin(x)}{x}$$
$$\frac{-1}{2}u'-u\frac{(1+x\sin(x))}{3x}= -\frac{\sin(x)}{x}$$
$$u'+u\frac{2(1+x\sin(x))}{3x}= \frac{2\sin(x)}{x}$$
The integrating factor and several steps become very messy, so it is probably wrong somewhere here. Is this the right approach?

Your approach is correct, but you have made a mistake. The correct differential equation would be:
$$y'-y\frac {1+x\sin x}{3x}=-\frac {y^{\mathbf 4} \sin x}{x}$$
This is indeed the Bernoulli form, so the procedure ahead is clear.
The IF would not come out to be messy or complicated, in fact it is of the form $x e^{-\cos x}$, which when multiplied with $\frac {\sin x}{x}$ gives an elementary integral solvable by substituting $t=\cos x$..
