We are asked to find $\frac{dy}{dx}$ for the equation:

$2x^3 - 5xy - y^2 = 3$

Since this equation involves both $x$ and $y$, we need to use implicit differentiation. Let's proceed step by step.

Step 1: Differentiate each term with respect to $x$

1. $\frac{d}{dx}(2x^3) = 6x^2$
2. $\frac{d}{dx}(-5xy)$ — This requires the product rule: $\frac{d}{dx}(-5xy) = -5\left(\frac{d}{dx}(x \cdot y)\right) = -5\left(1 \cdot y + x \cdot \frac{dy}{dx}\right) = -5y - 5x \frac{dy}{dx}$
3. $\frac{d}{dx}(-y^2) = -2y \frac{dy}{dx}$ — Since $y$ is a function of $x$, we apply the chain rule.

Step 2: Differentiate the right-hand side

$\frac{d}{dx}(3) = 0$

Step 3: Combine the results

After differentiating each term, we get:

$6x^2 - 5y - 5x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0$

Step 4: Solve for $\frac{dy}{dx}$

Now, let's solve for $\frac{dy}{dx}$:

$6x^2 - 5y = 5x \frac{dy}{dx} + 2y \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the right-hand side:

$6x^2 - 5y = \left(5x + 2y\right) \frac{dy}{dx}$

Finally, divide both sides by $5x + 2y$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{6x^2 - 5y}{5x + 2y}$

This is the derivative $\frac{dy}{dx}$.

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Would you like more details on any part of this process or have any further questions?

Here are 5 related questions to expand on this topic:

1. What is the product rule, and how does it apply in implicit differentiation?
2. Can you show how to solve another implicit differentiation problem with a trigonometric function?
3. How do you apply the chain rule when differentiating composite functions?
4. Why do we need implicit differentiation when differentiating functions involving both $x$ and $y$?
5. How would the solution change if the original equation was $2x^3 - 5xy + y^2 = 3$?

Tip: When differentiating terms like $xy$, always apply the product rule, since both $x$ and $y$ are functions of $x$.