Euler's identity to find integrating factor for an homogeneous 1-form This question is probably very elementary but I don't know how to tackle the conversely part of the following result. Let $M(x,y)$ and $N(x,y)$ be two differentiable and homogeneous functions of the same degree $d$ and such that $M(x,y)dx+N(x,y)dy$ is not exact that is:
$$ \frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x} $$
Using Euler's identity on $M$ and $N$: 
$$ x\cdot M_{x}(x,y)+y\cdot M_{y}(x,y)=d\cdot M(x,y)$$
$$ x\cdot N_{x}(x,y)+y\cdot N_{y}(x,y)=d\cdot N(x,y)$$
Euler's identity are true for each because $M$ and $N$ are homogenoeus functions of the same degree $d$.
I showed that the function
$$\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)}$$
will satisfy:
$$ \frac{\partial}{\partial y}\left(\,\mu\cdot M\right)=\frac{\partial}{\partial x}\left(\mu\cdot N\right).$$
My question: If we suppose that the PDE above is true how we can show that one solution for $\mu(x,y)$ is the fraction given above. That is:
$${If}\quad N(x,y)\mu_{x}-M(x,y)\mu_{y}=(N_{x}-M_{y})\mu,\quad \text{then where the formula} \quad\mu(x,y)=\frac{1}{xM(x,y)+yN(x,y)} \quad \text{comes from ?} $$
I tried to apply the method of characteristics but I don't see it. This result comes from an old edition of Boyce and DiPrima.

(From Serret J.A. Cours de Calcul Differentiel Et Integral... Volume 1 book)  Consider the homogeneous 1-form:
$$ M(x,y)dx+N(x,y)dy=0 $$
Since $M(x,y)$ and $N(x,y)$ are both homogeneous functions of the same degree we can find a one variable function $f$ such that:
$$ \frac{M(x,y)}{N(x,y)}=f\left(\frac{y}{x}\right), $$
then our original form becomes:
$$ N(x,y)\bigg(\frac{M(x,y)}{N(x,y)}dx+dy\bigg)=0 $$
$$ N(x,y)\bigg(f\left(\frac{y}{x}\right)dx+dy\bigg)=0 $$
We know that an homogeneous differential 1-form can be turned into a separable differential equation using the change of variables 
$$z=\frac{y}{x}$$
and this give us:
$$ N(x,zx)\bigg(f(z)dx+d(zx)\bigg)=0 $$
then
$$ N(x,zx)\bigg(f(z)dx+xdz+zdx\bigg)=0 $$
$$ N(x,zx)\bigg(\left(f(z)+z\right)dx+xdz\bigg)=0 $$
$$ xN(x,zx)\big(f(z)+z\big)\left(\frac{dx}{x}+\frac{dz}{f(z)+z}\right)=0$$
From here we see that multiplying the last equation by $\displaystyle{\frac{1}{xN(x,zx)\big(f(z)+z\big)}=\frac{1}{xM(x,y)+yN(x,y)}}$ give us the following equation:
$$ \frac{dx}{x}+\frac{dz}{f(z)+z}=0 $$
which is exact because:
$$ \frac{\partial}{\partial z}\left(\frac{1}{x}\right)=0=\frac{\partial}{\partial x}\left(\frac{1}{f(z)+z}\right)$$
Remark: $f(z)+z\neq 0$ because $\displaystyle{\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}}.$
