Solve of the differential equation $\left(3y^2+x^2+x+2y+1\right)\cdot y'+2xy+y=0$ I have some problem. 
There is an equation:
$$\left(3y^2+x^2+x+2y+1\right)\cdot y'+2xy+y=0$$
Open brackets. 
$$3y^2dy+x^2dy+2xdx y+xdy +ydx +2ydy+dy=0$$
But what to do, tell me, please?
I saw this a duplicate:
Find the general solution of the differential equation $\left(3y^2+x^2+x+2y+1\right)\cdot y'+2xy+y=0$
But he resolved properly, I ask to prompt more options.
 A: Anyway here's a solution, since you are asking for intermediate steps. 
$3y^2dy+x^2dy+2xdx y+xdy +ydx +2ydy+dy=0$
$\color{blue}{\overbrace{3y^2dy+2ydy+dy}}+\color{red}{\underbrace{x^2dy +2xdxy}} + \color{green}{\overbrace{xdy + ydx}}=0$
Notice the expression coloured, They can be written as 
$\color{blue}{3y^2dy+2ydy+dy=d(y^3+y^2+y)}$
$\color{red}{x^2dy +2xdxy=d(x^2y)}$
$\color{green}{xdy + ydx=d(xy)}$
Substituting and using linearity of derivative. 
$d(y^3+y^2+y+x^2y+xy)=0$
Now integrate.
A: I do not understand how this is different but in general it is extremely useful to be able to recognize these things:
$$y^2y'=\frac{(y^3)'}{3}$$
$$yy'=\frac{(y^2)'}{2}$$
$$xy'+y=(xy)'$$
These come up all the time.
A: ops . a simple way to solve this problem! this Differential equation is exact!
$(3y^2+x^2+x+2y+1)dy+(2xy+y)dx=0$
$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}=2x+1$
$\frac{\partial f(x,y)}{\partial y}=N(x,y)$
$f(x,y)={\int (3y^2+x^2+x+2y+1)dy}+h(x)$
$f(x,y)=y^3+yx^2+xy+y^2+y+h(x)$
$\frac{\partial f(x,y)}{\partial x}=M(x,y)$
$2xy+y+h'(x)=2xy+y$
$h'(x)=0$ , $h(x)=c$
$$f(x,y)=y^3+yx^2+xy+y^2+y=C$$
