Two masses m_1 and m_2 are placed on a frictionless horizontal...

Since no external force acts on the system in the horizontal direction, the center of mass remains at rest.

Let $d_1$ and $d_2$ be the distances moved by $m_1$ and $m_2$ respectively before they come to rest again.

From the conservation of the center of mass position:

$m_1 d_1 = m_2 d_2$

The system is released from rest with the spring stretched by $x_0$. The masses will come to rest again at the other extreme position of the oscillation, where the spring is compressed by $x_0$.

The total change in the length of the spring is $x_0 - (-x_0) = 2x_0$.

Since the masses move towards each other, the sum of the distances moved by them is equal to the total change in the length of the spring:

$d_1 + d_2 = 2x_0$

Substituting $d_2 = \dfrac{m_1 d_1}{m_2}$ into the above equation:

$d_1 + \dfrac{m_1 d_1}{m_2} = 2x_0$

$d_1 \left( \dfrac{m_1 + m_2}{m_2} \right) = 2x_0$

$d_1 = \dfrac{2m_2 x_0}{m_1 + m_2}$

Similarly, the distance moved by $m_2$ is:

$d_2 = \dfrac{2m_1 x_0}{m_1 + m_2}$