Non-Exact solution to ODE Question:
$$ \text{Solve:}  \; \; (3xy+y^2)dx+(x^2+xy)dy=0$$
Solution Thus far:
$$ M = 3xy+y^2 \; \text{and} \; N= x^2+xy $$
$$ \frac{\partial{M(x,y)}}{\partial{y}} = 3x + 2y$$
$$ \frac{\partial{N(x,y)}}{\partial{x}} = 2x + y$$
$$ \implies \frac{\partial{N}}{\partial{x}} -\frac{\partial{M}}{\partial{y}} \neq 0 $$
Hence this is not an exact equation therefore we must apply an integrating factor such that:
$$ \mu(x,y)M (x,y)+ \mu(x,y) N (x,y) = 0 $$ which is an exact equation
, meaning this is equivalent so solving the equation:
$$ \frac{\partial{\mu(x,y)M(x,y)}}{\partial{y}} = \frac{\partial{\mu(x,y)N(x,y)}}{\partial{x}} $$
Now rearrange this into a more computationally friendly form:
$$ (-N(x,y)) \frac{\partial{\mu}}{\partial{x}} + M(x,y)\frac{\partial{\mu}}{\partial{y}} =  \bigg{(} \frac{\partial{N}}{\partial{x}} -\frac{\partial{M}}{\partial{y}} \bigg{)} \mu $$
$ \text{define:} \; \mu = \text{integrating factor} $
Where I'm stuck:
$ \mu (x,y) = \mu(x) $ did not work
$ \mu (x,y) = \mu(y) $ did not work
$ \mu (x,y) = \mu(x+y) $ nearly worked but didn't
This guess and testing of an integrating factor is too difficult for me. How does one gain the intuition to find it. Or am I using the wrong technique entirely? Any insight would be greatly appreciated!

We are given
$$(3xy+y^2)dx+(x^2+xy)dy=0$$
Rearrange the differential equation to
$$\frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}$$
then let $$y=vx,\quad \frac{dy}{dx}=v + x\frac{dv}{dx}$$ 
The differential equation becomes
$$v + x\frac{dv}{dx}=-\frac{3x^2v+v^2x^2}{x^2+x^2v}$$
$$v + x\frac{dv}{dx}=-\frac{3v+v^2}{1+v}$$
$$x\frac{dv}{dx}=-\frac{4v+2v^2}{1+v}$$
$$\frac{1+v}{4v+2v^2}\,dv=-\frac{1}{x}\,dx$$
$$\frac{2+2v}{2v+v^2}\,dv=-\frac{4}{x}\,dx$$
Integrate both sides. 
In general, a common strategy is to rewrite the first order differential equation as a homogeneous equation of the form
$$\frac{dy}{dx}=f(x,y)$$
from which a common substitution is $y=vx$. The first order equation can then be solved as a separable equation after you substitute the values of $y$ and $dy$ and then simplify.
