Integrating factor: DE $ f(xy)ydx+g(xy)xdy=0 $ Let the DE be of the following form:
$ f(xy)ydx+g(xy)xdy=0 $ 
If $xy(f(xy)-g(xy)) \neq 0$ show that $\mu(x,y)=\frac 1 {xy(f(xy)-g(xy))}$ is this DEs integrating factor.
EDIT: Please note its $f(xy) , g(xy)$ and not $f(x,y), g(x,y)$
EDIT#2: 
I set $f(xy)x=P$ and $g(xy)y=Q$ 
$(\mu P)_y = (\mu Q)_x$
$f\mu + y f_y\mu+ yf\mu_y = g\mu + xf_x\mu+xf\mu_x$
deviding by $\mu$
$\frac {\mu_y} \mu yf +f_y y + f = \frac {\mu_x} \mu xg + g_x\mu + g$
$\frac {\mu_y} \mu = \frac {-x(f-g)-xy(f_y-g_y)}{xy(f-g)}=\frac{-1}y - \frac{f_y - g_y}{(f-g)}$ analogue we get  $\frac {\mu_x}\mu=\frac{-1}x - \frac{f_x - g_x}{(f-g)}$
Now if I carry on i get to 
$x(fg_x-f_xg)=y(fg_y-f_yg)$ and I don't know what to do here actually quite stuck. And wondering if I am missing an easyer path.
EDIT#3: 
While writing this noticed that $f=f(xy) , g=g(xy)$
Could it be that I missed it.
$\frac 1 y f_x=\frac 1 x f_y$ and same for $g(xy)$.
Expressing the last solution I got to $f(yg_y-xg_x)=g(yf_y-xf_x)$
Making the solution trivial once we enter the newly found out $\frac 1 y f_x=\frac 1 x f_y$ and $\frac 1 y g_x=\frac 1 x g_y$.
NB: $f_y = \frac{\delta}{\delta y} f(xy)$
 A: For the DEQ of the form:
$$\tag 1 x~ M~dx + y~N~dy = y~ f(xy)~dx + x~ g(xy)~dy=0 $$ 
Show that $\dfrac{1}{M x - N y}$, for $M x - N y \neq 0$, an integrating factor is given by
$$\tag 2 \mu(x,y)=\dfrac 1 {xy(f(xy)-g(xy))}$$
Multiplying $(1)$ by $(2)$
$\tag 3 \dfrac{y f(xy)}{xy(f(xy)-g(xy))}~dx+\dfrac{x g(x, y)}{xy(f(xy)-g(xy))}~dy   = 0$
Now we have to show that $(3)$ is exact.
Step 1: For $M = f [x(f-g)]^{-1}$, we use the Product Rule and have
$$\tag 4 M_y = f_y [x(f-g)]^{-1} -f[x(f-g)]^{-2}x(f_y-g_y) =\dfrac{f g_y-g f_y }{x(f-g)^2 }$$
Step 2: For $N = g[y(f-g)]^{-1}$, we have
$$\tag 5 N_x = g_x [y(f-g)]^{-1} -g[y(f-g)]^{-2}y(f_x-g_x)=\dfrac{f g_x -g f_x}{y(f-g)^2 }$$
Step 3: Find $M_y - N_x$ as
$$M_y - N_x = \dfrac{f g_y-g f_y }{x(f-g)^2 } -\dfrac{f g_x -g f_x}{y(f-g)^2 }  = \dfrac{f(y g_y - x g_x)+g(-yf_y+xf_x)}{xy (f-g)^2 }$$
The last result is identically zero because
$$y \dfrac{\partial g(xy)}{\partial y} = x\dfrac{\partial g(xy)}{\partial x} \\ y \dfrac{\partial f(xy)}{\partial y} = x\dfrac{\partial f(xy)}{\partial x}$$
$$\frac{f*0+g*0}{xy(f-g)^2}=0$$
