Problem regarding unique solution of differential equation 
A unique  solution  to the  differential  equation  $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist

then choose the  correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 , y_0)$
My attempt  : Here $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$
Now i put  $x= e^z$ then $z= \log x$
So $y= Dy- D^2y$ 
$D^2y-Dy -y=0$
so $(D^2-D-1)y=0$
so auxiliary equation will  be $m^2-m-1=0$ ,$m= \frac{1 +_{-}\sqrt - 3}{2}$
so $y= e^{\frac{1}{2}x} (c_1 \cos(\frac{\sqrt - 3}{2} ) + c_2\sin ({\frac{\sqrt - 3}{2}} )x)$
After that im not able  to proceed further

$$(y')^2-xy'=-y$$
Complete the square:
$$(y')^2-xy'+\frac {x^2} 4 =-y+\frac {x^2} 4 $$
$$(y'-\frac x 2 )^2=-y+\dfrac {x^2} 4 $$
The differential equation reduces to:
$$(z')^2=z$$
Where $z=-y+\dfrac {x^2} 4 $
Now you have to analyse what happens when $z <0,z=0,z>0$
