Differential equation. $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ Check my steps I have equation $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ and I need to solve it with separation method.
My try:
$-x^2\cdot dy-x\cdot dy=-\sqrt{3+y^2}\cdot dx$
$-x(x+1)\cdot dy=-\sqrt{3+y^2}\cdot dx $
$\frac{-x(x+1)}{dx}=-\frac{\sqrt{3+y^2}}{dy}$
And another step should be integrals for both sides. Are my steps correct or not?

Just rewrite the equation as
$$
\dfrac{1}{x+x^2} dx = \frac{1}{\sqrt{3+y^2}}dy.
$$
By integration, you get
$$
\log \frac{|x|}{|1+x|} = \textrm{arcsinh} y + C 
$$
So, the solution can be given in an explicit form:
$$
y = \sinh\left(\log \dfrac{|x|}{|x+1|}-C\right).
$$
