The answer is yes. The shorter diagonal divides the rhombus into two equilateral triangles with the sides of length $60$ cm. Each of the two triangles can be further divided into $4$ equilateral triangles with the sides of length $30$ cm. We have thus divided the rhombus into $8$ equilateral triangles with the sides measuring $30$ cm. Since there are $9$ points inside the rhombus, at least one of these equilateral triangles must contain at least two points. Let us prove that two points contained in the same triangle are at most $30$ cm apart. Denote the points by $E$ and $F$ and denote the vertices of the equilateral triangle containing them by $A$, $B$ and $C$. Consider the triangle $AEF$. The inner angle at $A$ measures $60^\circ$ or less, so one of the remaining two angles must measure at least $60^\circ$. From here we conclude that $EF$ is not the longest side in the triangle $AEF$. The other two sides measure at most $30$ cm since the points $E$ and $F$ lie inside a circle centered at $A$ with radius $30$ cm. Thus, the distance between $E$ and $F$ is at most $30$ cm.