How should I apply the homogeneous condition to check $\mu$ to be an integral factor? Assume that
$$
M(x,y)dx+N(x,y)dy=0
$$
is a homogeneous equation, and that $xM+yN \neq0$. Verify that
$$
\mu=\frac{1}{xM+yN}
$$
is an integral factor.

I've done some calculation:
$$
\frac{\partial(\mu M)}{\partial y}=\frac{-x\frac{\partial{M}}{\partial{y}}-N-y\frac{\partial{N}}{\partial{y}}}{(xM+yN)^{2}}M+\mu\frac{\partial{M}}{\partial{y}} \\
 \frac{\partial(\mu N)}{\partial x}=\frac{-M-x\frac{\partial{M}}{\partial{x}}-y\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}}N+\mu\frac{\partial{N}}{\partial{x}} \\
\frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x}=\frac{xN\frac{\partial{M}}{\partial{x}}-yM\frac{\partial{N}}{\partial{y}}+yN\frac{\partial{M}}{\partial{y}}-xM\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}}
$$
I want to prove that $\frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x}=0$, but got stuck. I think I cannot go furthur since I don't know how to apply the homogeneous condition here. Thank you in advance.
 A: $$M(x,y)dx+N(x,y)dy=0$$
Note that:
$$y=tx \implies dy=xdt+tdx$$
$$(M+Nt)dx+Nxdt=0$$
Multiply by integrating factor:
$$\dfrac {dx}x+\dfrac {N}{M+Nt}dt=0$$
Note that since $M,N$ are homogeneous functions:
$$\dfrac {dx}x+\dfrac {N(x,y)}{M(x,y)+tN(x,y)}dt=0$$
$$\dfrac {dx}x+\dfrac {N(1,t)}{M(1,t)+tN(1,t)}dt=0$$
$$\implies f(x)dx+g(t)dt=0$$
This is integrable:
$$\implies F(x)+G(t)=C$$
Note that you are asked to check the given integrating factor $\mu$ not to find it.

$N$ is homogeneous so we can do this:
$$N(x,y)=N(x,tx)=xN(1,t)$$
Do the same for the function $M$. I considered the functions as of degree $1$ of homogeneity. But it's nearly the same for another degree.
A: As I have commented
\begin{align} \frac{\partial(\mu M)}{\partial y}-\frac{\partial(\mu N)}{\partial x} &=\frac{xN\frac{\partial{M}}{\partial{x}}-yM\frac{\partial{N}}{\partial{y}}+yN\frac{\partial{M}}{\partial{y}}-xM\frac{\partial{N}}{\partial{x}}}{(xM+yN)^{2}}\\
&= \frac{N^2}{(xM+yN)^{2}}( x\ \partial_x + y\ \partial_y)\frac{M}{N} \\
&= \frac{N^2}{(xM+yN)^{2}}(\mathbf x \cdot\nabla)\frac{M}{N}
\end{align}
Euler's homogeneous function theorem states for a homogeneous function of degree $k$ we have
$$
\mathbf x \cdot\nabla f(\mathbf x) = kf(\mathbf x)
$$
Since both the functions $M$ and $N$ are assumed to be of the same degree (by defintion of the homogeneous equation), then their quotient is homogeneous of degree zero (check it!). It follows that
$$
\mathbf x \cdot\nabla\frac{M}{N}(\mathbf x) = 0
$$
