If $2x + 3y$ is multiple of $17$, then $9x + 5y$ is multiple of $17$
Solution 1:
$$3(9x+5y)-5(2x+3y) =17x$$
Since $3$and $17$ are relatively prime if $2x+3y$ Is a multiple of $17$ so is $9x+5y$
Solution 2:
Your proof is not correct since $9$ and $5$ never really come into play.
But you may notice that $\frac{9}{2}=9(2)^{-1}$ and $\frac{5}{3}=5(3)^{-1}$ are the same class $\!\!\pmod{17}$, namely $13$.
So $2x+3y\equiv 0\pmod{17}$ implies $13(2x+3y)\equiv 0\pmod{17}$, i.e. $9x+5y\equiv 0\pmod{17}$.