Differential equation (Clairaut) Problem: I have to solve the following differential equation (which is supposed to be Clairaut's):
$$y=x(y')^2-\frac{1}{y'},$$
where $y'=\frac{dy}{dx}.$
What I have tried: Because it says that is Clairaut's, I proceed by differentiating with respect to $x$. With an ideal CLairaut's differential equation, you could "isolate" every term that is multiplied by $\frac{d^2y}{dx}$ and, on the other side of the equality, you will get a zero. From there, it easy and you end up getting a family of curves [Link: Wikipedia]. The problem is the square in $x(y')^2,$ that messes up everything and I cannot isolate the terms in the right way.
I also tried using Mathematica 
DSolve[{y[x] == x (y[x]')^2 - 1/(y[x]')}, y[x], x],

but it gives an error 
$RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of -(1/y').

I tried Python with Sympy, but failed too.
I tried a variable change, but I do not manage to find the optimal to simplify the equation and solve it following the regular procedure.

This is a d'Alembert's equation.  Letting $p=y'$, we have
$$
y=xp^2-\frac1p
$$
Differentiating with respect to $x$,
$$
p=p^2+2xp\frac{\mathrm{d}p}{\mathrm{d}x}+\frac1{p^2}\frac{\mathrm{d}p}{\mathrm{d}x}
$$
So
$$
\frac{\mathrm{d}x}{\mathrm{d}p}=\frac{2}{1-p}x+\frac1{p^3-p^4}
$$
which integrates to $x = \frac{c}{(p-1)^2} - \frac1{2 (p - 1)^2 p^2} + \frac1{(p - 1)^2 p}$ and thus
$$
y=xp^2-\frac1p=\dots
$$
