In the usual coordinate system $ xOy$ a line $ d$ intersect the circles $ C_1:$ $ (x\plus{}1)^2\plus{}y^2\equal{}1$ and $ C_2:$ $ (x\minus{}2)^2\plus{}y^2\equal{}4$ in the points $ A,B,C$ and $ D$ (in this order). It is known that $ A\left(\minus{}\frac32,\frac{\sqrt3}2\right)$ and $ \angle{BOC}\equal{}60^{\circ}$. All the $ Oy$ coordinates of these $ 4$ points are positive. Find the slope of $ d$.
1. **Identify the given information and the equations of the circles:**
   - Circle \( C_1 \): \((x + 1)^2 + y^2 = 1\)
   - Circle \( C_2 \): \((x - 2)^2 + y^2 = 4\)
   - Point \( A \left( -\frac{3}{2}, \frac{\sqrt{3}}{2} \right) \) lies on circle \( C_1 \).
   - \(\angle BOC = 60^\circ\)
   - All \( y \)-coordinates of points \( A, B, C, D \) are positive.

2. **Verify that point \( A \) lies on circle \( C_1 \):**
   \[
   \left( -\frac{3}{2} + 1 \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 = \left( -\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{1}{4} + \frac{3}{4} = 1
   \]
   This confirms that point \( A \) lies on circle \( C_1 \).

3. **Determine the coordinates of points \( B \) and \( C \):**
   - Since \( \angle BOC = 60^\circ \), points \( B \) and \( C \) must lie on circle \( C_2 \) and form an equilateral triangle with the origin \( O \).

4. **Find the coordinates of points \( B \) and \( C \):**
   - The center of circle \( C_2 \) is at \( (2, 0) \) with radius \( 2 \).
   - Since \( \angle BOC = 60^\circ \), the points \( B \) and \( C \) are at \( 60^\circ \) and \( 120^\circ \) from the center of circle \( C_2 \).

5. **Calculate the coordinates of points \( B \) and \( C \):**
   - Point \( B \) at \( 60^\circ \) from the center \( (2, 0) \):
     \[
     B = \left( 2 + 2 \cos 60^\circ, 2 \sin 60^\circ \right) = \left( 2 + 2 \cdot \frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2} \right) = (3, \sqrt{3})
     \]
   - Point \( C \) at \( 120^\circ \) from the center \( (2, 0) \):
     \[
     C = \left( 2 + 2 \cos 120^\circ, 2 \sin 120^\circ \right) = \left( 2 + 2 \cdot \left( -\frac{1}{2} \right), 2 \cdot \frac{\sqrt{3}}{2} \right) = (1, \sqrt{3})
     \]

6. **Determine the coordinates of point \( D \):**
   - Since \( D \) is the fourth intersection point and all \( y \)-coordinates are positive, \( D \) must lie on circle \( C_2 \) and be symmetric to \( B \) with respect to the \( y \)-axis.
   - Therefore, \( D = (3, \sqrt{3}) \).

7. **Find the slope of line \( d \):**
   - Line \( d \) passes through points \( A \left( -\frac{3}{2}, \frac{\sqrt{3}}{2} \right) \) and \( D (3, \sqrt{3}) \).
   - The slope \( m \) of line \( d \) is given by:
     \[
     m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{3} - \frac{\sqrt{3}}{2}}{3 - \left( -\frac{3}{2} \right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{9}{2}} = \frac{\sqrt{3}}{9}
     \]

The final answer is \(\boxed{\frac{\sqrt{3}}{9}}\).