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To determine the interval where the function \( f(x) = x^3 \ln(x) \) is concave down, we need to follow these steps:

### Step 1: Compute the First and Second Derivatives
To find where a function is concave down, we need to find the second derivative, \( f''(x) \), and analyze where it is negative.

1. **First derivative** \( f'(x) \):
   \[
   f'(x) = 3x^2 \ln(x) + x^2 \cdot \frac{1}{x} = 3x^2 \ln(x) + x
   \]

2. **Second derivative** \( f''(x) \):
   Apply the product rule and chain rule:
   \[
   f''(x) = \frac{d}{dx}(3x^2 \ln(x)) + \frac{d}{dx}(x)
   \]
   \[
   f''(x) = 6x \ln(x) + 3x + 1
   \]
   So,
   \[
   f''(x) = 6x \ln(x) + 4x
   \]

### Step 2: Solve \( f''(x) = 0 \) to Find Inflection Points
To find the critical points where the concavity changes, set \( f''(x) = 0 \):
\[
6x \ln(x) + 4x = 0
\]
Factor out \( x \):
\[
x(6 \ln(x) + 4) = 0
\]
Since \( x = 0 \) is not in the domain (as \( \ln(x) \) is undefined at 0), we solve:
\[
6 \ln(x) + 4 = 0
\]
\[
\ln(x) = -\frac{2}{3}
\]
\[
x = e^{-2/3}
\]

### Step 3: Test Intervals Around \( x = e^{-2/3} \)
We now test the sign of \( f''(x) \) on the intervals \( (0, e^{-2/3}) \) and \( (e^{-2/3}, \infty) \).

- For \( x > e^{-2/3} \), the term \( 6 \ln(x) \) becomes positive, making \( f''(x) > 0 \), which indicates concave **up**.
- For \( 0 < x < e^{-2/3} \), the term \( 6 \ln(x) \) is negative, making \( f''(x) < 0 \), which indicates concave **down**.

### Conclusion:
The function \( f(x) = x^3 \ln(x) \) is concave down on the interval \( (0, e^{-2/3}) \).

Thus, the correct answer is **(D)**.

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Do you need any more details or clarifications? Here are some related questions you might want to explore:

1. How do you interpret concavity in terms of the behavior of a function?
2. What is the significance of the second derivative in determining concavity?
3. How would you find the inflection points for more complex functions?
4. Can concavity change more than once on the domain of a function?
5. How does the logarithmic function affect the overall concavity?

**Tip**: When solving for concavity, always check the second derivative's sign on either side of critical points to ensure accurate interval classification.

Let f(x) = x^3 ln x. On what interval is f concave down?

Find the interval where the function f(x) = x^3 ln x is concave down by computing its second derivative and solving for critical points. This problem applies key calculus concepts such as derivatives and the concavity test.

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