A first order non-linear ordinary differential equation How do we solve the first order non-linear ordinary differential equation , 
$$y^3=xy^2 \frac{dy}{dx} + x^4 \Big(\frac{dy}{dx}\Big)^2$$ 

Firstly we note that: $$y(x)=0 \tag{Solution 1}$$ is a solution, hereafter we take $y(x)\ne0$.
Substitute: $$y(x)=\dfrac{1}{h(x)}$$ into:$$ y \left( x \right)   ^{3}=x  y \left( x \right) 
  ^{2}{\frac {d}{dx}}y \left( x \right) +{x}^{4} \left( {\frac 
{d}{dx}}y \left( x \right)  \right) ^{2}\tag{1}$$
to get:
$$\begin{align}
h \left( x \right) &=-x{\frac {d}{dx}}h \left( x \right) +{x}^{4}
 \left( {\frac {d}{dx}}h \left( x \right)  \right) ^{2} \\
h \left( x \right) +\dfrac{1}{4x^2}&={x}^{4} \left( {\frac {d}{dx}}h
 \left( x \right) -\dfrac{1}{2x^3} \right) ^{2}\\ 
&={x}^{4} \left( {\frac {d}{dx}}\left(h
 \left( x \right) +\dfrac{1}{4x^2}\right) \right) ^{2}\tag{2}
\end{align}$$
and we see that:
\begin{align} h(x)&=-\dfrac{1}{4x^2}\\
y(x)&=-4x^2 \tag{Solution 2}\end{align}
is a solution (as found by @Daniel Littlewood and @user64494). Note: in $(2)$ we divided by $1/h(x)^4$ under the assumption it was not zero everywhere, if it were we would recover $\text{Solution 1}$. Then substitute:
$$h(x)=g \left( x \right) -\dfrac{1}{4x^2}$$
into $(2)$ to get:
$$g \left( x \right) ={x}^{4} \left( {\frac {d}{dx}}g \left( x \right) 
 \right) ^{2}\tag{3}$$
and change the variable to $x=\dfrac{1}{\xi}$ in $(3)$ and write $g(x)=f(x)^2$ to get:
\begin{align}g \left( \xi \right) &=\left( {\frac {d}{d\xi}}g \left( \xi \right) 
 \right) ^{2}\\ 
f\left( \xi \right)^2 &=4f\left( \xi \right)^2\left( {\frac {d}{d\xi}}f\left( \xi \right) 
 \right) ^{2}\tag{4}\end{align}
we then divide by $f(\xi)^2$ under the assumption it is not zero everywhere, if it were we would recover $\text{Solution 2}$, and we have:
\begin{align} {\frac {d}{d\xi}}f \left( \xi \right)&=\pm\dfrac{1}{2}\\
f \left( \xi \right)&=C\pm\dfrac{\xi}{2} \tag{5}\end{align}
but by noting that $g=f^2$ we can encompass either sign possibility in the arbitrary constant $C$ and write:
$$\begin{align}
g(\xi)&=\left(C+\dfrac{\xi}{2}\right)^2 \\
g(x)&=\left(C+\dfrac{1}{2x}\right)^2\tag{6}
\end{align}$$
Working backwards we then have:
$$y \left( x \right) ={\frac {1}{\left(\dfrac{C}{2}\right)^2 +\left(\dfrac{C}{2}\right)\dfrac{1}{x}}}. \tag{7}$$
To fix the constant in $(7)$ we find that the starting value is useless because, irrespective of $C$:
$$\lim_{x\rightarrow0}y(x)=0$$
but we can use the starting value of the derivative (Neumann boundary condition) to find that:
$$C=\dfrac{2}{y'(0)}$$
$$y \left( x \right) ={\frac {y'(0)^2}{1 +y'(0)\dfrac{1}{x}}}={\frac {xy'(0)^2}{x +y'(0)}} \tag{Solution 3}$$
so this solution is bilinear in $x$. We also note that $\text{Solution 1}$ is encompassed in $\text{Solution 3}$ but $\text{Solution 2}$ is distinct.
