Solve 2nd order Ordinary differential equation $y d^2y/dx^2 -2 (dy/dx)^2 = y^2$ $$y \dfrac{d^2y}{dx^2} -2\left( \dfrac{dy}{dx} \right)^2 =y^2$$
let $y \dfrac{dy}{dx} = t$
$$\left( \dfrac{dy}{dx}) \right)^2 +y\dfrac{d^2y}{dx^2}=\dfrac{dt}{dx}$$
put in equation
after this i am stuck I don't how to proceed because I got differential in $x ,y$, and $t$  please help 
 A: Noting the form of the LHS, it is instructive to choose the following path:
$${yy’’-2{y’}^2\over y^2} = 1$$
$$\frac{d\frac{y’}y}{dx} = 1+{\biggl(\frac {y’}y\biggr)}^2$$
$$x+c = \arctan{\frac{y’}y}$$
$$y’-y\tan{(x+c)}=0$$
To get the solution for $(x,y)$ this linear D. E. must be solved. For some values of $x$, as mentioned in the comments by @WE, there is indeed a solution.
$$y’(\cos {(x+c)})-y(\sin {(x+c)}) = 0$$
$$(y\cos{(x+c)})’ = 0$$
$$\implies y\cos{(x+c)} = k$$
$$\boxed{y = ksec{(x+c)}}$$
A: Starting from Certainly not a dog's answer
$$y’-y\tan{(x+c)}=0\implies \frac{y'}y=\tan{(x+c)}$$ Integrate both sides to get
$$\log(y)=-\log (\cos (x+c))+d\implies y=d \sec(x+c)$$
A: Hint.
From
$$
y y''-2(y')^2=y^2\Rightarrow \frac{y''}{y}-2\left(\frac{y'}{y}\right)^2=1
$$
but
$$
\left(\frac{y'}{y}\right)' = -\left(\frac{y'}{y}\right)^2+\frac{y''}{y}
$$
then
$$
\left(\frac{y'}{y}\right)'-\left(\frac{y'}{y}\right)^2 = 1
$$
now calling $\zeta = \frac{y'}{y}$ we have
$$
\zeta'-\zeta^2 = 1
$$
which is separable.
