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 first change from $y'$ to $x'$, then get a reduction of order.  The inverse function formula for the second derivative is $y'' = -\left(\frac{1}{x'}\right)^3x''$.  Therefore, your equation becomes:
$$-y\,\left(\frac{1}{x'}\right)^3x'' = 3\left(\frac{1}{x'}\right)^2$$
This reduces to:
$$-y\,\left(\frac{1}{x'}\right)x'' = 3$$
Now you have a problem where you can do reduction of order for $x$.  So, $u = x'$ and $u' = x''$.
$$ -y\,\frac{1}{u}\,u' = 3 $$
Alternatively, you can write $u'$ as $\frac{du}{dy}$.  So, this becomes:
$$ -y\,\frac{1}{u}\,\frac{du}{dy} = 3 \\
 \frac{du}{u} = -3\frac{dy}{y} \\
\ln(u) = -3\,\ln(y) + C \\
u = \frac{C}{y^3} \\
\frac{dx}{dy} = \frac{C}{y^3} \\
dx = C y^{-3}\,dy \\
x = Cy^{-2} + D 
 $$
Note that a few times in there, C got merged with another constant, but I didn't feel like spelling it out.
