To solve for $y''$ in the equation provided:

$\frac{d}{dx}(2x - 6y - 6xy' + 3y^2y') = \frac{d}{dx} 0$

we proceed as follows:

1. Since the right side is the derivative of 0 with respect to $x$, which equals 0, we only need to focus on differentiating the left side with respect to $x$.

2. Expanding the left side:

   $\frac{d}{dx}(2x - 6y - 6xy' + 3y^2y') = 0$

3. Now, differentiate each term on the left side with respect to $x$:

   • First term: $\frac{d}{dx}(2x) = 2$
   • Second term: $\frac{d}{dx}(-6y) = -6y'$
   • Third term: $\frac{d}{dx}(-6xy') = -6 \left( y' + xy'' \right)$, using the product rule
   • Fourth term: $\frac{d}{dx}(3y^2y') = 3(2y y'y' + y^2 y'') = 6yy'^2 + 3y^2y''$, using the product rule and chain rule

4. Combine these differentiated terms:

   $2 - 6y' - 6(y' + xy'') + 6yy'^2 + 3y^2y'' = 0$

5. Now, distribute and collect terms involving $y''$:

   $2 - 6y' - 6y' - 6xy'' + 6yy'^2 + 3y^2y'' = 0$

6. Combine like terms:

   $2 - 12y' + 6yy'^2 + (3y^2 - 6x)y'' = 0$

7. Solving for $y''$:

   $(3y^2 - 6x)y'' = -2 + 12y' - 6yy'^2$

   $y'' = \frac{-2 + 12y' - 6yy'^2}{3y^2 - 6x}$

Thus, the solution for $y''$ is:

$y'' = \frac{-2 + 12y' - 6yy'^2}{3y^2 - 6x}$

Would you like further details on any step, or do you have any questions?

Here are some related questions to deepen understanding:

1. How would we solve for $y'$ if this equation were implicit?
2. What would happen if we included an initial condition for $y$ and $y'$?
3. How do the product and chain rules work in similar derivative problems?
4. How could we apply a similar approach to find higher-order derivatives?
5. What are the implications if $3y^2 - 6x = 0$ for the solution?

Tip: Always carefully apply the product rule when differentiating terms involving products of functions, especially when one term itself is a derivative.