Suppose we have $Mdx+ Ndy=0$ We want to transform it into variable u&x such that $x=uy$
My attempt is below!
$$\frac{dy}{dx}=g(1/u)$$
$$\frac{dy}{dx}=\frac{u-x\frac{du}{dx}}{u^2}$$
After getting the new separable differential equation use it to solve
$$(x^2-3y^2)dx+2xydy=0$$
So given $\bbox[lemonchiffon,0.25ex]{x=uy}$ , then $\bbox[lemonchiffon,0.25ex]{\mathrm dx=y\,\mathrm du + u\,\mathrm dy}$, by the product rule. Rearranging:
$$\mathrm d y = u^{-1}\,\mathrm d x-u^{-1}y\,\mathrm d u$$
Tip: $y$ terms are undesirable, so we substitute: $y\gets u^{-1}x$.
$$\mathrm d y = u^{-1}\,\mathrm d x-u^{-2}x\,\mathrm d u$$
Then using this to substitute into $\bbox[lemonchiffon,0.25ex]{N\mathrm d x+M\mathrm d y=0}$ we find...