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The problem asks to calculate the second derivative \( y'' = \frac{d^2 y}{dx^2} \) for the function \( y = y(x) \), where the relationship between \( x \) and \( y \) is given by parametric equations:

\[
\begin{cases}
x = \ln t \\
y = t^3
\end{cases}
\]

### Solution Steps:

1. **Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)**:
   - From \( y = t^3 \), we differentiate with respect to \( t \):
     \[
     \frac{dy}{dt} = 3t^2
     \]
   - From \( x = \ln t \), we differentiate with respect to \( t \):
     \[
     \frac{dx}{dt} = \frac{1}{t}
     \]

2. **Find the first derivative \(\frac{dy}{dx}\)**:
   Using the chain rule:
   \[
   \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{\frac{1}{t}} = 3t^3
   \]

3. **Find the second derivative \(\frac{d^2 y}{dx^2}\)**:
   To find the second derivative, first differentiate \(\frac{dy}{dx} = 3t^3\) with respect to \( t \):
   \[
   \frac{d}{dt}(3t^3) = 9t^2
   \]
   Now, divide this result by \(\frac{dx}{dt} = \frac{1}{t}\) to get the second derivative:
   \[
   \frac{d^2 y}{dx^2} = \frac{9t^2}{\frac{1}{t}} = 9t^3
   \]

Thus, the second derivative is:

\[
y'' = 9t^3
\]

If you'd like to express this in terms of \( x \), recall that \( t = e^x \) since \( x = \ln t \). Substituting \( t = e^x \) into the expression for \( y'' \):

\[
y'' = 9(e^x)^3 = 9e^{3x}
\]

### Final Answer:
\[
y'' = 9e^{3x}
\]

Let me know if you need further clarification or details.

Here are some related questions:
1. How do parametric equations work in general?
2. How is the chain rule used in parametric differentiation?
3. Can you find the curvature of a curve using second derivatives?
4. What are the applications of parametric equations in physics?
5. How can implicit differentiation be used for non-parametric forms?

**Tip:** In parametric differentiation, always differentiate both the numerator and denominator with respect to the parameter before dividing.

Tính đạo hàm y'' = y''(x) của hàm số y = y(x) cho bởi phương trình tham số { x = ln(t), y = t^3 }.

Learn how to calculate the second derivative y'' for the parametric equations x = ln(t) and y = t^3. This solution demonstrates the use of the chain rule and parametric differentiation. Suitable for calculus students.

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