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The given problem asks for the nature of the point \( z = 0 \) for the function:

\[
f(z) = \frac{z - \sin(\sin(z))}{z^3}.
\]

### Step 1: Identifying potential issues at \( z = 0 \)
The function is expressed as a fraction, and we are evaluating the behavior around \( z = 0 \). At \( z = 0 \), the denominator is 0, which suggests a singularity or an issue in defining the function directly. To classify the nature of the singularity, we need to analyze the behavior of both the numerator and denominator near \( z = 0 \).

---

### Step 2: Expanding the numerator using Taylor series
For small \( z \), we can approximate \( \sin(z) \) and \( \sin(\sin(z)) \) using their Taylor series expansions:
1. **Taylor expansion of \( \sin(z) \):**
   \[
   \sin(z) = z - \frac{z^3}{6} + \mathcal{O}(z^5).
   \]
2. **Taylor expansion of \( \sin(\sin(z)) \):**
   Substituting the first few terms of \( \sin(z) \) into itself:
   \[
   \sin(\sin(z)) = \sin\left(z - \frac{z^3}{6} + \mathcal{O}(z^5)\right) = z - \frac{z^3}{6} + \mathcal{O}(z^5).
   \]

Thus:
\[
z - \sin(\sin(z)) = z - \left(z - \frac{z^3}{6} + \mathcal{O}(z^5)\right) = \frac{z^3}{6} + \mathcal{O}(z^5).
\]

---

### Step 3: Simplifying the function \( f(z) \)
We now substitute the above expansion into the function:

\[
f(z) = \frac{\frac{z^3}{6} + \mathcal{O}(z^5)}{z^3} = \frac{1}{6} + \mathcal{O}(z^2).
\]

As \( z \to 0 \), the function \( f(z) \) approaches \( \frac{1}{6} \), which means it has a **finite limit** at \( z = 0 \).

---

### Step 4: Classifying the singularity
Since the function \( f(z) \) has a finite value at \( z = 0 \), the singularity at this point is **removable**. That is, we can redefine the function at \( z = 0 \) to be \( f(0) = \frac{1}{6} \), making the function continuous.

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### Final Answer
The correct classification for \( z = 0 \) is:

**Removable singularity.**

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Would you like further details or additional examples of different types of singularities?

### Related Questions:
1. What is the difference between a pole and a removable singularity?
2. How do you compute Laurent series for complex functions?
3. What are essential singularities and how do they differ from poles?
4. Can a singularity be both isolated and removable?
5. How is the behavior of a function near a singularity related to its residue?

**Tip:** Removable singularities occur when the limit of a function exists at a point where it is not originally defined.

If f(z) = (z - sin(sin(z))) / z^3, then z = 0 is:

Analyze the singularity at z = 0 for the function f(z) = (z - sin(sin(z))) / z^3. This problem involves complex analysis concepts like singularities and Taylor series expansions, and concludes that z = 0 is a removable singularity.

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