How many hexagons can be made by joining the vertices of a 15 sided polygon if none of the sides of the hexagon is also the side of the 15-gon.

The $15$-sided polygon is a fixed object. You need a reference point. Say it is vertex $A$.

There are two possibilities: vertex $A$ is a vertex of the hexagon, or it is not.

Suppose $A$ is not a vertex of the hexagon. Arrange nine blue balls in the form of a circle. Label one of those balls $A$. Choose six of the nine gaps in which to place a green ball to ensure that no two green balls are adjacent. Now label the vertices in alphabetical order as you proceed clockwise around the circle, starting at $A$. The letters on the green balls are the vertices of the hexagon. There are $\binom{9}{6}$ ways to do this, which agrees with your answer.

Now suppose $A$ is a vertex of the hexagon. Color it green. Arrange nine blue balls around $A$ to form a circle which includes the ball labeled $A$. Doing so creates $10$ spaces in which to place a green ball, but two of these are adjacent to $A$, leaving eight spaces where a green ball could be placed so that it is not adjacent to $A$. Choose five of these eight spaces in which to place a green ball to ensure that no two green balls are adjacent. Now label the balls in alphabetical order as you proceed clockwise around the circle, starting at $A$. The letters on the green balls are the vertices of the hexagon. There are $\binom{8}{5}$ ways to do this, which you have not taken into account.

Since the two cases are mutually exclusive and exhaustive, the number of admissible hexagons is $$\binom{9}{6} + \binom{8}{5}$$ which agrees with the linked answers provided by Andre Nicolas and Henry.

What you are missing is a multiplication factor of $\frac{15}{15-6}$

ie the answer should be $\binom96\times \frac{15}{9}$

The multiplication factor is arising because you are necessarily attaching a gap (clockwise, say) immediately after each vertex, which restricts the number of places the vertex-gap "blocks" can occupy.