Proving that a function $\mu (x,y)$ is an integrating factor of a first order homogeneous ODE. The question is from Differential Equations by SL Ross Book:

Show that if the equation $$M(x,y)dx +N(x,y)dy=0 \tag{A}\label{A}$$ is homogeneous and $M(x,y)x +N(x,y)y \neq 0$, then $\frac1{[M(x,y)x +N(x,y)y]}$ is an integrating factor of $(\ref{A})$

$\textbf{Attempt 1}$
I started with writing $(\ref A)$ in derivative form $$ \frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}=g\left(\frac{y}x\right) \text{  (say)  }$$
Now, multiplying both sides of equation $(\ref A)$ by $\frac1{[M(x,y)x +N(x,y)y]}=\mu(x,y) \text{ (say)}\:(\because M(x,y)x +N(x,y)y \neq 0 \text{ ,given})$
we get,
${M \over Mx+Ny}dx+{N \over Mx+Ny}dy=0 \tag{i}\label{i}$
Let $M_1(x,y)={M \over Mx+Ny}, N_1(x,y)={N \over Mx+Ny}$
Thus, $\mu(x,y)$ will be an integrating factor of $(\ref A)$ iff equation $(\ref i)$ is exact, i.e., ${\partial M_1 \over \partial y}={\partial N_1 \over \partial x}$
It can be easily noticed that,
$\begin{align} M_1&= \frac{M}{N} N_1 \; (\because N(x,y) \neq 0)\\
\implies M_1&=-g\left(\frac{y}x\right) N_1 \end{align}$
I thought this relation would prove quite useful, but then
$\begin{align}{\partial M_1 \over \partial y}&=-N_1{\partial  \over \partial y}\left(g\left(\frac{y}x\right)\right)-g\left(\frac{y}x\right) {\partial N_1 \over \partial y}\end{align}$
which I need to prove equal to ${\partial N_1 \over \partial x}$. But when I couldn't proceed further, I decided not to use that relation:
$\textbf{Attempt 2}$
$\begin{align}{\partial M_1 \over \partial y}&={\partial \over \partial y}\left({M \over Mx+Ny}\right)\\&={(Mx+Ny){\partial M \over \partial y}-M{\partial \over \partial y} (Mx+Ny) \over (Mx+Ny)^2}\\&={(Mx+Ny){\partial M \over \partial y}-M[x{\partial M \over \partial y}+y{\partial N \over \partial y}+N] \over (Mx+Ny)^2}\\&={Ny{\partial M \over \partial y}-My{\partial N \over \partial y}-MN \over (Mx+Ny)^2}\end{align} $
And,
$\begin{align}{\partial N_1 \over \partial x}&={\partial \over \partial x}\left({N \over Mx+Ny}\right)\\&={(Mx+Ny){\partial N \over \partial x}-N{\partial \over \partial x} (Mx+Ny) \over (Mx+Ny)^2}\\&={(Mx+Ny){\partial N \over \partial x}-N[x{\partial M \over \partial x}+M+y{\partial N \over \partial x}] \over (Mx+Ny)^2}\\&={Mx{\partial N \over \partial x}-Nx{\partial M \over \partial x}-MN \over (Mx+Ny)^2} \end{align}$
I haven't used the fact that equation (A) is homogeneous in Attempt 2 yet and don't know where or in what way should I use that. Using either methods,  I haven't been able to show ${\partial M_1 \over \partial y}={\partial N_1 \over \partial x}$
I think I am missing something because the question seems quite simple. Any hint or a solution using any other method will be appreciated.
Note:
(i) About equation (A) being "homogeneous". It has nothing to do with that $n^{th}$ order linear ordinary "homogeneous" equation definition which is simply a linear ODE with right hand member $b(x)=0$. Here homogeneous equation (first order homogeneous equation, they mean) means the differential equation if, when written in derivative form $\frac{dy}{dx}=f(x,y),\; \exists$ a function, $g$ such that, $f(x,y)=g\left(\frac{y}x\right)$. I don't understand why they use the same name everywhere creating confusion.
(ii) This question: A formula for homogeneous differential form integrating factor
does not answer my question. There, the OP asked about finding the IF while I am asking to verify the same IF (which he/she had already done but not posted anything related to that).
 A: Functions $M_1$ and $N_1$ are homogeneous:
$$
M_1(\lambda x, \lambda y) = \frac1\lambda M_1(x, y), \quad \forall \lambda, \quad(0)
$$
Hence we have
$$
M_1 + x\frac{\partial M_1}{\partial x} + y\frac{\partial M_1}{\partial y} = 0 \quad (1)
$$
By definition
$$
xM_1 + yN_1 = 1.
$$
Differentiating last equality with respect to x, we obtain
$$
M_1 + x\frac{\partial M_1}{\partial x} + y\frac{\partial N_1}{\partial x} = 0 \quad (2)
$$
Equalities (1) and (2) are valid for all $x, y$ only if
$$
\frac{\partial M_1}{\partial y} = \frac{\partial N_1}{\partial x}.
$$
Update. To obtain (1) it is enough to differentiate (0) with respect to $\lambda$ and take $\lambda = 1$.
A: From equation (0) to equation (1) in Gec's answer
$M_1(\lambda x, \lambda y) = \frac1\lambda M_1(x, y), \quad \forall \lambda, \quad(0)$
Partially differentiating the above equation w.r.t  $\lambda$
$ \begin{align} & \frac{\partial M_1(\lambda x, \lambda y)}{\partial \lambda}= -\frac{M_1(x,y)}{{\lambda}^2}\\
 \implies & \frac{\partial M_1(\lambda x, \lambda y)}{\partial \lambda x} \frac{\partial \lambda x}{\partial \lambda} + \frac{\partial M_1(\lambda x, \lambda y)}{\partial \lambda y} \frac{\partial \lambda y}{\partial \lambda}=-\frac{M_1}{{\lambda}^2} & \text{  (Using Chain Rule)}\\
 \implies & x \frac{\partial M_1(\lambda x, \lambda y)}{\partial \lambda x}  + y \frac{\partial M_1(\lambda x, \lambda y)}{\partial \lambda y}+ \frac{M_1}{{\lambda}^2}=0 \\
\implies & x \frac{\partial M_1(x, y)}{\partial x}  + y \frac{\partial M_1(x,y)}{\partial y}+ M_1=0 & \text{  (for $\lambda=1$)} \\
\implies & M_1 + x\frac{\partial M_1}{\partial x} + y\frac{\partial M_1}{\partial y} = 0 \quad (1) \end{align}$
