is this exact?
2x/y^3 dx     +       y^2 - 3x^2 / y^4  dy     = 0

Let's analyze the equation:

\[
\frac{2x}{y^3} \, dx + \left( \frac{y^2 - 3x^2}{y^4} \right) dy = 0
\]

This is indeed a **differential equation**. However, for clarity:

1. The term \(\frac{y^2 - 3x^2}{y^4}\) simplifies further:
   \[
   \frac{y^2}{y^4} - \frac{3x^2}{y^4} = \frac{1}{y^2} - \frac{3x^2}{y^4}
   \]

Thus, the equation becomes:
\[
\frac{2x}{y^3} \, dx + \left( \frac{1}{y^2} - \frac{3x^2}{y^4} \right) dy = 0
\]

Now it's in its simplified form. Does this answer your question? Would you like me to help with solving or further analyzing this equation?

Here are 5 related questions to explore:
1. What methods can be used to solve this type of differential equation?
2. How do you solve first-order differential equations in exact form?
3. What is the role of integrating factors in differential equations?
4. How can we test if a differential equation is exact?
5. How do we handle nonlinear terms in differential equations?

**Tip:** Always simplify terms before attempting to solve differential equations for easier manipulation.

Explore how to simplify and solve the differential equation 2x/y^3 dx + (y^2 - 3x^2)/y^4 dy = 0 using techniques such as simplification and exactness tests. This solution breaks down complex fractions and provides step-by-step instructions on how to approach first-order differential equations. Suitable for advanced calculus students.

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