The given expression appears to be a first-order differential equation. It can be written as:

$3y' + \frac{3}{x}y^3 = 0$

We can rewrite it in a more standard form:

$y' + \frac{1}{x}y^3 = 0$

This is a separable differential equation. Let's solve it step-by-step:

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Step 1: Rewrite in separable form

Move all terms involving $y$ to one side and those involving $x$ to the other:

$y' = -\frac{1}{x}y^3$

Express $y'$ as $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{1}{x}y^3$

Rearrange to separate variables:

$\frac{dy}{y^3} = -\frac{dx}{x}$

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Step 2: Integrate both sides

The left-hand side integrates as:

$\int y^{-3} \, dy = \int -\frac{1}{x} \, dx$

For the left-hand side:

$\int y^{-3} \, dy = \frac{-1}{2}y^{-2}$

For the right-hand side:

$\int -\frac{1}{x} \, dx = -\ln|x|$

Thus, the equation becomes:

$\frac{-1}{2}y^{-2} = -\ln|x| + C$

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Step 3: Simplify the solution

Multiply through by $-2$ to eliminate the negative fraction:

$y^{-2} = 2\ln|x| - 2C$

Let $C_1 = -2C$, so:

$y^{-2} = 2\ln|x| + C_1$

Rewrite in terms of $y$:

$y = \pm \frac{1}{\sqrt{2\ln|x| + C_1}}$

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Final Solution:

The solution is:

$y = \pm \frac{1}{\sqrt{2\ln|x| + C_1}}$

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Would you like additional clarification or details about this solution? Let me know!

Related Questions:

1. How do separable differential equations differ from linear equations?
2. What other techniques are used for solving first-order differential equations?
3. Can you explain the role of constants of integration in differential equations?
4. How is the natural logarithm function used in solving differential equations?
5. What happens when initial conditions are given for such an equation?

Tip: Always check whether a differential equation is separable; it simplifies solving dramatically!