Help me understand this solution $$2x^4yy'+y^4=4x^6$$
The way my teacher did it is:
First, he made a substitution:
$y=z^m$
$y'=mz^{m-1}z'$
$$2x^4 z^m mz^{m-1} z'+z^{4m}=4x^6$$
$$z'=\frac{4x^6-z^{4m}}{2mx^4z^{2m-1}}=f\left(\frac{z}{x}\right)$$
Then he wrote the part I don't understand (how he got $m$):
$2m-1=2$
$2m+1=4$
$m=\frac{3}{2}$ - for this $m$, the equation becomes homogeneous.
So we make a substitution: $y=z^{\frac{3}{2}}$
$$y'=\frac{3}{2}z^{\frac{1}{2}} z'$$
$$3x^4z^2z'+z^6=4x^6$$
Substitution: $z(x)=u(x) x$
$$\frac{u^2}{u-u^6-3u^3}du=\frac{dx}{3x}$$
We multiply by $-3$:
$$\frac{3u^2}{u^6+3u^3-u}du=-\frac{dx}{x}$$
Substitution: $u^3=t$, $3u^2\,du=dt$
$$\frac{x^3-y^2}{4x^3-y^2}=\frac{c1}{x^5}$$
$c1=c$ or $c1=-c$
So, all in all, the part I don't understand is how he got m.

here ia nother way to do this problem. 
a change of variable $y = u^k$ in $$2x^4 yy'+y^4 = 4x^6  \to 2kx^4u^{2k-1}u'+u^{4k} = 4x^6$$ now we will choose $k$ so that $2k-1 = 4k \to k = -1/2.$ and we have $$-x^4u'+u= 4x^6 \to e^{x^3/3}u' - \frac{e^{x^3/3}}{x^4}u = -4x^6e^{x^3/3} $$ on integration, we get $$u =-4e^{-x^3/3}\int 4x^6e^{x^3/3}\, dx, \quad y = \frac1{u^2}.$$
