Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$ Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$
My Attempt:
$$p^3x-p^2y-1=0$$
$$p^2y=p^3x-1$$
$$y=px-\frac {1}{p^2}$$
This is solvable for $y$ so differentiating both sides w.r.t $x$
$$\frac {dy}{dx} = p + x \cdot \frac {dp}{dx} - (-2) p^{-3} \cdot \frac {dp}{dx}$$
$$p=p+x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}$$
$$x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}=0$$
$$(x+\frac {2}{p^3}) \cdot \frac {dp}{dx}=0$$
How do I solve further?

The initial equation is
$$y'x-y-\frac1{y'^2}=0$$
and by differentiating,
$$y''x+2\frac{y''}{y'^3}=0$$
We split in two cases:
$$y''=0\to y=ax+b$$
and
$$y'=-\sqrt[3]{\frac2x}\to y=-\frac32\sqrt[3]{2x^2}+c.$$

Plugging back in the equation
$$p=a\to a^3x-a^2(ax+b)-1=0$$ implies $b=-\dfrac1{a^2}$, and 
$$p=-\sqrt[3]{\frac2x}\to-\frac2xx+\sqrt[3]{\frac4{x^2}}\left(\frac32\sqrt[3]{2x^2}+c\right)-1=0,$$ 
requires $c=0.$

Finally 
$$y=ax-\frac1{a^2}$$ or $$y=-\frac32\sqrt[3]{2x^2}.$$
