Find the derivative of the solution of the equation $\ddot{x} = \dot{x}^2 + x^3$ with initial condition $x(0) = 0$, $\dot{x}(0) = A$ with respect to $A$ for $A = 0$.
1. **Given the differential equation:**
   \[
   \ddot{x} = \dot{x}^2 + x^3
   \]
   with initial conditions \( x(0) = 0 \) and \( \dot{x}(0) = A \).

2. **Differentiate the given equation with respect to \( A \):**
   \[
   \frac{d}{dA}(\ddot{x}) = \frac{d}{dA}(\dot{x}^2 + x^3)
   \]
   Using the chain rule, we get:
   \[
   \frac{d}{dA}(\ddot{x}) = \frac{d\ddot{x}}{dA} = \frac{d}{dA}(\dot{x}^2) + \frac{d}{dA}(x^3)
   \]
   \[
   \frac{d\ddot{x}}{dA} = 2\dot{x} \frac{d\dot{x}}{dA} + 3x^2 \frac{dx}{dA}
   \]

3. **Evaluate the expression at \( A = 0 \):**
   Given \( x(0) = 0 \) and \( \dot{x}(0) = A \), when \( A = 0 \), we have \( x = 0 \) and \( \dot{x} = 0 \). Thus:
   \[
   \frac{d\ddot{x}}{dA}\bigg|_{A=0} = 2 \cdot 0 \cdot \frac{d\dot{x}}{dA}\bigg|_{A=0} + 3 \cdot 0^2 \cdot \frac{dx}{dA}\bigg|_{A=0} = 0
   \]

4. **Define \( y = \frac{dx}{dA}\bigg|_{A=0} \):**
   Then, \( y \) satisfies the differential equation:
   \[
   \ddot{y} = 0
   \]

5. **Compute the initial values for \( y \):**
   Since \( x(0) = 0 \) and \( \dot{x}(0) = A \), we have:
   \[
   y(0) = \frac{dx}{dA}\bigg|_{A=0, t=0} = 0
   \]
   \[
   \dot{y}(0) = \frac{d\dot{x}}{dA}\bigg|_{A=0, t=0} = 1
   \]

6. **Solve the differential equation \( \ddot{y} = 0 \) with the initial conditions:**
   The general solution to \( \ddot{y} = 0 \) is:
   \[
   y(t) = C_1 t + C_2
   \]
   Using the initial conditions \( y(0) = 0 \) and \( \dot{y}(0) = 1 \):
   \[
   y(0) = C_2 = 0 \implies C_2 = 0
   \]
   \[
   \dot{y}(0) = C_1 = 1 \implies C_1 = 1
   \]
   Therefore, the solution is:
   \[
   y(t) = t
   \]

7. **Thus, we have:**
   \[
   \frac{dx}{dA}\bigg|_{A=0} = t
   \]

The final answer is \(\boxed{t}\).