To solve for $y''$ (the second derivative of $y$ with respect to $x$) of the equation $x^3 - y^3 = 3$, we will need to implicitly differentiate both sides of the equation.

Given:

$x^3 - y^3 = 3$

Step 1: Differentiate implicitly with respect to $x$.

Differentiating both sides of the equation with respect to $x$, we treat $y$ as a function of $x$ (i.e., $y = y(x)$):

$\frac{d}{dx}(x^3) - \frac{d}{dx}(y^3) = \frac{d}{dx}(3)$

This gives:

$3x^2 - 3y^2 \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$.

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

$3x^2 = 3y^2 \frac{dy}{dx}$

Divide both sides by $3y^2$:

$\frac{dy}{dx} = \frac{x^2}{y^2}$

Step 3: Differentiate again to find $y''$.

Now, differentiate $\frac{dy}{dx} = \frac{x^2}{y^2}$ implicitly to get the second derivative $y''$:

$\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{x^2}{y^2} \right)$

Using the quotient rule on the right-hand side:

$\frac{d}{dx} \left( \frac{x^2}{y^2} \right) = \frac{2x y^2 - x^2 \cdot 2y \frac{dy}{dx}}{y^4}$

Now substitute $\frac{dy}{dx} = \frac{x^2}{y^2}$ into this:

$y'' = \frac{2x y^2 - x^2 \cdot 2y \cdot \frac{x^2}{y^2}}{y^4}$

Simplify the expression:

$y'' = \frac{2x y^2 - 2x^4 y / y^2}{y^4}$

$y'' = \frac{2x y^2 - 2x^4 / y}{y^4}$

This is the second derivative $y''$.

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Do you need further details or clarification on any of the steps? Here are some related questions you might find interesting:

1. How can we apply implicit differentiation in other types of equations?
2. What is the quotient rule and how is it used in differentiation?
3. Can this method be applied to higher-order derivatives?
4. How do you solve for $y'$ and $y''$ in parametric equations?
5. What other methods exist for solving implicit differentiation problems?

Tip: Always simplify the expression at each step to avoid unnecessary complexity.