$ AA_{3}$ and $ BB_{3}$ are altitudes of acute-angled $ \triangle ABC$. Points $ A_{1}$ and $ B_{1}$ are second points of intersection lines $ AA_{3}$ and $ BB_{3}$ with circumcircle of $ \triangle ABC$ respectively. $ A_{2}$ and $ B_{2}$ are points on $ BC$ and $ AC$ respectively. $ A_{1}A_{2}\parallel AC$, $ B_{1}B_{2}\parallel BC$. Point $ M$ is midpoint of $ A_{2}B_{2}$. $ \angle BCA \equal{} x$. Find $ \angle A_{3}MB_{3}$.
1. **Identify the orthocenter and cyclic quadrilaterals:**
   Let \( H \) be the orthocenter of \( \triangle ABC \). The points \( A_1 \) and \( B_1 \) are the second intersections of the altitudes \( AA_3 \) and \( BB_3 \) with the circumcircle of \( \triangle ABC \), respectively. Since \( A_1A_2 \parallel AC \) and \( B_1B_2 \parallel BC \), the points \( A_2 \) and \( B_2 \) lie on \( BC \) and \( AC \), respectively.

2. **Define the intersection points:**
   Let \( D \) be the intersection of \( A_1A_2 \) and \( B_1B_2 \). The lines \( A_1A_2 \) and \( B_1B_2 \) are parallel to \( AC \) and \( BC \), respectively. Let \( E \) and \( F \) be the intersections of \( B_1B_2 \) with \( AA_3A_1 \) and \( A_1A_2 \) with \( BB_3B_1 \), respectively.

3. **Establish the cyclic nature of quadrilaterals:**
   The quadrilateral \( A_1B_1EF \) is cyclic because \( \angle A_1EB_1 = \angle A_1FB_1 = 90^\circ \). Since \( A_3 \) and \( B_3 \) are the midpoints of \( HA_1 \) and \( HB_1 \), respectively, we have:
   \[
   HA_3 \cdot HD = HB_3 \cdot HE
   \]
   This implies that the quadrilateral \( A_3B_3DE \) is also cyclic.

4. **Identify the circumcircle and its properties:**
   Let \( \mathcal{M} \) be the circumcircle of \( A_3B_3DE \). Let \( U \) and \( V \) be the other ends of the diameters spanning the right angles \( \angle A_1EB_1 \) and \( \angle A_1FB_1 \), respectively. Thus, \( \angle A_3EU = 90^\circ \) and \( \angle B_3FV = 90^\circ \).

5. **Analyze the parallelogram and its diagonals:**
   The quadrilateral \( A_3B_3UV \) is a parallelogram, and its diagonals intersect at the center of the circle \( \mathcal{M} \). This parallelogram is inscribed in the parallelogram \( CA_2DB_2 \), so they share the same diagonal intersection point \( M \), which is the midpoint of \( A_2B_2 \).

6. **Calculate the angle \( \angle A_3MB_3 \):**
   Since \( \triangle A_1FH \) and \( \triangle B_1EH \) are right-angled, and \( A_3 \) and \( B_3 \) are the midpoints of their hypotenuses \( HA_1 \) and \( HB_1 \), respectively, we have:
   \[
   \frac{\angle A_3MB_3}{2} = \angle A_3EB_3 = \angle HEB_3 = 90^\circ - \angle HB_1B_2 = 90^\circ - \angle A_1AC = \angle BCA = x
   \]
   Therefore,
   \[
   \angle A_3MB_3 = 2x
   \]

The final answer is \( \boxed{2x} \).