Solve $(x^2y^2-y)dx+(x+2x^3y)dy=0$ 
solve the following ODE $$(x^2y^2-y)dx+(x+2x^3y)dy=0$$ 

First I check exactness, $$M=x^2y^2-y,\quad N=x+2x^3y$$
$$\frac{\partial M}{\partial y}=2x^2y-1,\quad \frac{\partial N}{\partial x}=1+6x^2y$$
$M_y\neq N_x$, hence the ODE is not exact. Then I tried to reduced it to exact DE where I followed,

$(1)$ If $M$ and $N$ are homogeneous then, IF$=\frac{1}{M_x+N_y};M_x+N_y\neq0$
$(2)$ If the DE can be written as $yf_1(x,y)dx+xf_2(x,y)dy=0$ then, IF$=\frac{1}{M_x+N_y};{M_x+N_y}\neq0$
$(3)$ If $\frac{M_y-N_x}{N}=f(x)\text{ or C}$ then, IF$=e^{\int f(x)dx}$
$(4)$ If $\frac{M_y-N_x}{N}=f(y)\text{ or C}$ then, IF$=e^{-\int f(y)dy}$
  Where IF=Integrating factor

But none of these are satisfied. Then how to approach$?$ My professor said that inspection maybe a good approach for this one. But I haven't found any. And How to inspect any ODE$($like this$)$ to guess what to do$?$
 A: $$(x^2y^2-y)dx+(x+2x^3y)dy=0$$
$$xdy-ydx +x^2(y^2dx+2xydy)=0$$
Divide by $x^2$
$$\frac {xdy-ydx}{x^2} +(y^2dx+2xydy)=0$$
$$ {d}\left (\frac {y}{x}\right )+d(y^2x)=0$$
Integrate
A: Multiply by $x^ay^b$, so we want to solve$$\begin{align}(x^{a+2}y^{b+2}-x^ay^{b+1})_y&=(x^{a+1}y^b+2x^{a+3}y^{b+1})_x\\\implies(b+2)x^{a+2}y^{b+1}-(b+1)x^ay^b&=(a+1)x^ay^b+2(a+3)x^{a+2}y^{b+1}\\\implies(b+2,\,-b-1)&=(2a+6,\,a+1)\\\implies(a,\,b)&=(-2,\,0).\end{align}$$So$$0=(y^2-y/x^2)dx+(1/x+2xy)dy=d(xy^2+y/x)\implies x^2y^2-Cx+y=0.$$
A: Rewriting the equation as:
$$x^0y^0((-1)ydx + (1)xdy) + x^2y^1((1)ydx + (2)xdy) = 0$$
There's another rule that says that if a differential equation can be represented in the following form 
$$x^ay^b(mydx + nxdy) + x^{a'}y^{b'}(m'ydx + n'xdy) = 0$$ 
Then, its integrating factor is given by $x^hy^k$, where $h$ and $k$ are given by 
$$\frac{a + h + 1}{m} = \frac{b + k + 1}{n}$$
and 
$$\frac{a' + h + 1}{m'} = \frac{b' + k + 1}{n'}$$ 
Therefore, compare the first two equations to get the values and solve the equation to get the values of $h$ and $k$, and hence the integrating factor.
