Let's start by simplifying the given equation and adjusting it according to your requirements.

The equation provided is:

$(3x^2 + 4xy + 3) + (6y^2 + 2x^2 - 4) y' = 0$

Step 1: Rearrange the equation in implicit form

We have:

$(3x^2 + 4xy + 3) + (6y^2 + 2x^2 - 4)y' = 0$

Separate the terms containing $y'$ and those without it:

$(3x^2 + 4xy + 3) + y'(6y^2 + 2x^2 - 4) = 0$

This is in implicit form already.

Step 2: Ensure the highest degree term containing $x$ has coefficient 1

The highest degree term containing $x$ is $2x^2$ (from the $y'$-dependent term). To ensure its coefficient is 1, divide the entire equation by 2:

$\frac{3x^2 + 4xy + 3}{2} + y'\left( \frac{6y^2 + 2x^2 - 4}{2} \right) = 0$

Simplifying:

$\frac{3x^2}{2} + 2xy + \frac{3}{2} + y' \left( 3y^2 + x^2 - 2 \right) = 0$

Step 3: Move constants to the right-hand side

Now, move the constant term $\frac{3}{2}$ to the right-hand side:

$\frac{3x^2}{2} + 2xy + y' \left( 3y^2 + x^2 - 2 \right) = -\frac{3}{2}$

Final Result

The equation is now in the desired form:

$\frac{3x^2}{2} + 2xy + y' \left( 3y^2 + x^2 - 2 \right) = -\frac{3}{2}$

This satisfies all three conditions.

Would you like further clarification or more details? Here are some follow-up questions:

1. How does dividing by a coefficient affect the terms of an equation?
2. What are the differences between explicit and implicit forms of equations?
3. Why is it necessary to ensure the highest degree term has a coefficient of 1?
4. Can you derive the explicit form of this equation from the implicit form?
5. How do constants on the right-hand side affect the solution of a differential equation?

Tip: Always check if dividing the entire equation is needed to simplify the terms, especially when dealing with differential equations.