A first order non-linear ordinary differential equation How do we solve the first order non-linear ordinary differential equation , 
$$y^3=xy^2 \frac{dy}{dx} + x^4 \Big(\frac{dy}{dx}\Big)^2$$ 
 A: 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.
A: I don't know about a general solution, but I've found a particular one:
If $y$ is a polynomial of degree $k$, then the LHS has degree $3k$, the first term on the RHS has degree $3k$ and the second term has degree $2k+2$. If the two sides are identical, we require $2k+2\le3k$, so $k\ge2$, with equality at $k=2$. If we check, say, $y=ax^{2}$, we get
$$a^{3}x^{6}=2a^{3}x^{6}+4a^{2}x^{6}$$
So $a^{3}=-4a^{2}$, giving $a=0$ (trivial) or $a=-4$, so $y=-4x^{2}$. Using Wolfram Alpha, I have verified that this is the only solution of degree $\le 4$.  
A: These are all the solutions of this ODE, obtained using the Maple computer algebra system, and they are all verifiable (I would like to post with format, but not having 10 points the system does not allow me to upload an image, so it goes here in ascii):
Your equation - written in Maple syntax:

ode := y(x)^3-x*y(x)^2*(diff(y(x), x))-x^4*(diff(y(x), x))^2 = 0:

Five explicit solutions:

sol := dsolve(ode, Lie); # use Lie symmetry methods

                         / (1/2)          \      2  
               2         \2      _C1 - 2 x/ x _C1   
    y(x) = -4 x , y(x) = -------------------------, 
                                /   2      2\       
                              2 \_C1  - 2 x /       

               / (1/2)          \      2  
               \2      _C1 + 2 x/ x _C1   
      y(x) = - -------------------------, 
                      /   2      2\       
                    2 \_C1  - 2 x /       

                   / (1/2)          2  \  
               2 x \2      _C1 - _C1  x/  
      y(x) = - -------------------------, 
                     2 /   2  2    \      
                  _C1  \_C1  x  - 2/      

                 / (1/2)          2  \
             2 x \2      _C1 + _C1  x/
      y(x) = -------------------------
                   2 /   2  2    \    
                _C1  \_C1  x  - 2/    

where _C1 is an arbitarry constant. These five solutions verify OK; in Maple you test that these ODE solutions cancel the given ODE in this way:

map(odetest, [sol], ode);

                    [0, 0, 0, 0, 0]

To understand from where are these five solutions coming, you can always solve the equation implicitly, and you see three solutions, two of which are quadratic in y

sol := dsolve(ode, Lie, implicit);

         2  

y(x) = -4 x , 
                                         (1/2)            
             2        /    2      2     \                 
         y(x)    y(x) \y(x)  + 4 x  y(x)/                 
2 y(x) + ----- + ----------------------------- - _C1 = 0, 
           2                   2                          
          x                   x                           

                               (1/2)          
            /    2      2     \               
 2     1    \y(x)  + 4 x  y(x)/               
---- + -- + ------------------------ - _C1 = 0
y(x)    2                 2                   
       x            y(x) x 

To understand why three solutions, note that the ODE is nonlinear in the highest derivative y' and so there are singular cases. In the Maple system you compute the essential singular cases in this way:

PDEtools:-casesplit(ode, singsol = essential);

                                3         2 / d      \    
                        2   y(x)  - x y(x)  |--- y(x)|  
              / d      \                    \ dx     /  
              |--- y(x)|  = --------------------------, 
              \ dx     /                 4              
                                        x               

                           2
                y(x) = -4 x 

The meaning of this result is, basically, that there is a singular solution 
                                   2
                           y = -4 x 
that is not a particular case of the solutions of y'^2 = (y^3 - x y' y^2)/(x^4); in turn the latter, quadratic in y' leads to two solutions, each of them quadratic in y, shown above as the output of dsolve(ode, implicit)
Edgardo S. Cheb-Terrab
