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The problem is asking to find \( (f^{-1})'(-3) \) given that \( f(x) = x^3 - 3x \).

To solve this, let's follow these steps:

1. **Understand the Problem:**
   To find \( (f^{-1})'(-3) \), we can use the formula:
   \[
   (f^{-1})'(y) = \frac{1}{f'(x)}
   \]
   where \( y = f(x) \). Here, we need \( f(x) = -3 \) to find the corresponding \( x \) and then use \( f'(x) \) for that \( x \) value.

2. **Find \( x \) such that \( f(x) = -3 \):**
   Given \( f(x) = x^3 - 3x \), we need to solve:
   \[
   x^3 - 3x = -3
   \]
   Rearrange this to:
   \[
   x^3 - 3x + 3 = 0
   \]
   Solving this equation for \( x \) will give us the values needed.

3. **Differentiate \( f(x) \):**
   The derivative of \( f(x) = x^3 - 3x \) is:
   \[
   f'(x) = 3x^2 - 3
   \]
4. **Evaluate \( f'(x) \) at the appropriate \( x \) value:**
   Once we find the correct \( x \) that satisfies \( f(x) = -3 \), we can substitute it into \( f'(x) \) to find \( (f^{-1})'(-3) \).

Let me know if you'd like a step-by-step solution for solving for \( x \) and finishing the calculations. 

Here are related questions to expand your understanding:

1. How do we find the inverse of a function algebraically?
2. Why does the formula \( (f^{-1})'(y) = \frac{1}{f'(x)} \) work for derivatives of inverse functions?
3. What techniques can be used to solve cubic equations like \( x^3 - 3x + 3 = 0 \)?
4. How do we know if a function has an inverse over its entire domain?
5. What is the significance of differentiability in finding inverses?

**Tip:** When working with derivatives of inverse functions, ensure that \( f(x) \) is one-to-one in the region you're analyzing, as this guarantees an inverse exists for that range.

Learn how to find (f^{-1})'(-3) given that f(x) = x^3 - 3x. This calculus problem involves concepts like differentiation and the inverse function theorem. Suitable for students in Grades 11-12.

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