$(xy+2xy\ln^2y+y\ln y)\text{d}x+(2x^2\ln y+x)\text{d}y=0$ Solve: (Hint: use $x\ln y=t$)$$(xy+2xy\ln^2y+y\ln y)\text{d}x+(2x^2\ln y+x)\text{d}y=0$$
My Work:
$$x\ln y=t, \text{ d}t=\ln y \text{ d}x+\frac{x}{y} \text{ d}y$$
$$(x+2x\ln^2y+\ln y)\text{ d}x+\left(\frac{2x^2}{y}\ln y+\frac{x}{y}\right)\text{d}y=0$$
$$(x+2x\ln^2y)\text{ d}x+\left(\frac{2x^2}{y}\ln y\right)\text{d}y+\text{d}t=0$$
$$(x+2t\ln y)\text{ d}x+\left(\frac{2x}{y}t\right)\text{d}y+\text{d}t=0$$
$$\left(\frac{x}{t}+2\ln y\right)\text{d}x+\left(2\frac{x}{y}\right)\text{d}y+\frac{\text{d}t}{t}=0$$
$$\frac{x}{t}\text{d}x+2\text{ d}t+\frac{\text{d}t}{t}=0$$
$$\frac{x}{t}\text{d}x=\left(-2-\frac{1}{t}\right)\text{d}t$$
$$\frac{x^2}{2}=-t^2-t+c$$
$$\frac{x^2}{2}=-x^2\ln^2y-x\ln y+c$$
1. Is my answer correct? 
2. How we could recognize that we should use $x\ln y=t$. If question didn't Hint? 
3. All the way that I did to solve this equation was weird for me (because for example we had $dx,dy,dt$ in line 3) and had not saw this way to solve differential equation before is There any other way to simplification?
 A: $$(xy+2xy\ln^2 y+y\ln y)dx+(2x^2\ln y+x)dy=0$$
$$(x+2x\ln^2 y+\ln y )+(2x^2\ln y+x)(\ln y)'=0$$
2. How we could recognize that we should use xlny=t. If question didn't Hint? 
Only $\ln y $  terms in the equation. Substitute $z=\ln y$
$$(x+2xz^2+z)+(2x^2z+x)z'=0$$
$$(x+2xz^2+z)+x^2(z^2)' +xz'=0$$
$$ x +(x^2z^2)' +(xz)'=0$$
Integrate and substitute back:
$$ \frac 12x^2 +x^2z^2 +xz=0$$
$$ \frac 12x^2 +x^2\ln^2y +x\ln y =K$$ 
A: I am using the product rule of differentiation in reverse
Divide throughout by $y$
$$\left(x+2x(lny)^2+lny\right)dx+\left(2x^2\frac{lny}{y}+\frac{x}{y}\right)dy=0$$
$$xdx+(lny)^2d(x^2)+x^2d((lny)^2)+(lny)d(x)+xd(lny)=0$$
$$xdx+d(x^2(lny)^2)+d(xlny)=0$$
Integrating
$$\frac{x^2}{2}+x^2(lny)^2+xlny=c$$
A: *

*You're right. Your final result implies $x=-2x\ln^2y-\frac{2x^2\ln y}{y}y^\prime-\ln y-\frac{x}{y}y^\prime$ and $y^\prime=-\frac{(2x\ln y+1)y\ln y+x}{x(1+2x\ln y)}$, which is equivalent to the original equation.

*The coefficients of $dx,\,dy$ are polynomials in $x,\,y,\,\ln y$, so an Ansatz $t=x^a\ln^by$ is natural. The result ends up especially simple if $a=b=1$.

*Given $f(x,\,y)dx+g(x,\,y)dy=0$, we hope some functions $h(x,\,y),\,j(x,\,y)$ satisfy $jf=\partial_xh,\,jg=\partial_yh$ (so that the equation is equivalent to $dh=0$), whence $\partial_x(jg)=\partial_y(jf)$. Again, there's a natural Ansatz, $j=x^cy^d\ln^ky$.

