Solve $y+xy'=x^4(y')^2$ The question is in the title, here $y'$ denotes the derivative of $y$ with respect to $x$. This doesn't appear to be in the scope of what I have learned so far. It is non-homogenous, non-linear, non-separable and non-exact equation. The LHS can easily be recognised as $\dfrac{d}{dx}(xy)$, but I cannot go further. 
Please help
 A: This looks similar to a Clairaut equation, so just try the derivative of this equation
$$
2y'+xy''=4x^3(y')^2+2x^4y'y''\iff 2y'(1-2x^3y')=xy''(-1+2x^3y')
$$
which means that there exists a nice factorization
$$
0=(2y'+xy'')(1-2x^3y')
$$
Each solution has to be zero in either the first or the second factor. The factor can change at points where both are simultaneously zero.
The first factor gives $y'=Cx^{-2}$ which can be inserted into the original equation to give $$y=-Cx^{-1}+C^2.$$
The second factor gives $y'=\frac1{2x^3}$ so that
$$
y=-\frac1{2x^2}+\frac{x^4}{4x^6}=-\frac1{4x^2}.
$$

To get the regular Clairaut form thus use $u=x^{-1}$ and $y(x)=v(u)=v(x^{-1})$, so that $y'(x)=-x^{-2}v'(x^{-1})$ giving
$$
v(u)-uv'(u)=v'(u)^2\iff v(u)=uv'(u)+v'(u)^2,
$$
which is a standard example.
A: $\newcommand{\d}{\mathrm{d}}$
$$\begin{align}\d y - z \d x &=0 &
y+z x-z^2x^4&=0\end{align}$$
$$\begin{split}
\d y + (x -2z x^4)\d z +(z - 4 z^2 x^3)\d x &= 0 \\
(x -2z x^4)\d z +(2z - 4 z^2 x^3)\d x &= 0 \\
(1 -2z x^3)(x \d z+2z \d x)&= 0
\end{split}$$
$$1-2zx^2=0\text{ or }x \d z + 2z \d x =0$$
and so on.
