Solve the DFE $xy'^3 - 2yy'^2 - 16x^2 = 0$ I've started like this:
From $xy'^3 - 2yy'^2 - 16x^2 = 0$, we find that $$y = -8x^2/p^2 + 1/2 \cdot xp,$$ with $p = y' = dy/dx$. 
Taking the derivative yields $$p = -16x/p^2 + 16 x^2p'/p^3 + 1/2p + 1/2 \cdot xp'$$ And working out the denominator gives us $$p^4 = -32xp + 32x^2 p' + xp^2p'.$$
At this point, I don't know how to go on. I tried separating the terms containing $p$ and $p'$, but it didn't get me anywhere. 
 A: We are given
$$\tag 1 xy'^3 - 2yy'^2 - 16x^2 = 0$$
We will rewrite $(1)$ using $p = \dfrac{dy}{dx}$ as
$$\tag 2 x p^3 - 2 y p^2 - 16 x^2 = 0$$
Isolating $y$ in $(2)$, we have
$$\tag 3 y = \dfrac{1}{2}( x p - 16 x^2 p^{-2})$$
Differentiating $(3$), we have
$$\tag 4 y ' = p = \dfrac{1}{2}(p + x p' - 32~ x~ p^{-2} + 32 ~x^2~ p^{-3}p')$$
Factoring $(4)$, we have
$$\tag 5 \dfrac{(32 x + p^3)(p - x p')}{2 p^3} = 0$$
From $(5)$, we solve for the two expressions
$$p^3 + 32 x = 0 \\ p - x p' = 0$$
I will assume you can take it from here.
A: $$p^4 = -32xp + 32x^2 p' + xp^2p'.$$
$$(32x^2 + xp^2)p'=p^4 +32xp  .$$
This is an generalized Abel's equation of the second kind :
$$y'\sum_{j=0}^{2}g_j(x)y^j=\sum_{j=0}^{4}f_j(x)y^j$$
where $g_0(x)=32x^2 \:;\:g_2(x)=x \:;\: f_1(x)=32x \:;\: f_4(x)=1 \:;\:g_1(x)=f_0(x)=f_2(x)=f_3(x)=0  .$ 
https://www.encyclopediaofmath.org/index.php/Abel_differential_equation
The majority of this kind of equations are not solvable on a closed form. Nevertheless, some particular solutions can sometimes be found. Knowing a boundary condition can eventually help.
In the case of the original equation :
$$xy'^3 - 2yy'^2 - 16x^2 = 0$$
looking for a particular solution on the form $y_p(x)=ax^p$ leads to :
$$y_p=-\frac{3}{\sqrt[3]{2}}x^{4/3}$$
