Solve the ODE $y y''=3(y')^2$ using Reduction of Order For the reduction of order method, we are supposed to guess a solution, $y_1$, and assume that the second solution (this will be second order as the title shows) is of the form $y_2=uy_1$.  But I'm having difficulty finding a solution by inspection.  I've tried the following, which seemed legit at first, only to find that the constant 3 is making the problem more difficult.
$$y=e^3x \quad y=e^{\sqrt{3}x} \quad y=ax^2 \quad y=\sqrt{x}$$
Previously the problems given, it was pretty straight forward but this one is really hurting as I can't find a solution by inspection.  It could be that we can maybe use some "product rule" manipulation, since
$$y y''=3(y')^2 \quad \rightarrow \quad y y''-3(y')^2=0$$
looks like it could be of the form $(y y')'$, but again, that 3 is messing me up.  Once I find it, I want to try and find the other solution using Reduction of Order.  So for anyone answering, please don't solve the ODE, just how I can maybe posit a guess given the information provided in the problem.

You can also write the d.e.
$$
\frac {y^{"}}{y^{'}}=3\frac{y^{'}}{y}\\
\frac{dy^{'}}{y^{'}}=3\frac{dy}{y}
$$
Integrate both sides,
$$
\log(\frac{y^{'}}{y^{'}_0})=3\log(\frac{y}{y_0}) \\
y^{'}=\frac{y^{'}_0}{y_0^{3}}y^3\\
$$
Integrate again,
$$
y=\frac{\gamma}{\sqrt{x_0-x}}\\
\gamma =\pm \sqrt{\frac{2y_0^3}{y^{'}_0}}
$$
