I assume $X_1$ and $X_2$ are independent.
First, here's some intuition: whenever you get an arrival in process $X_1$ (rate 1), your value of $Y$ goes down by 3. Whenever you get an arrival in process $X_2$ (rate 2), your value goes up by 2. Therefore, we could build a merged poisson process with rate 3, and for each arrival we obtain value -3 with probability 1/3 and value 2 with probability 2/3. Our total value $Y$ is the sum of all the rewards collected at each poisson arrival.
Formalizing, if we define $N\sim \text{Pois}(\lambda_1 + \lambda_2)$, $M_n\sim\left\{\begin{array}{ll} -3 & w.p.~\lambda_1/(\lambda_1+\lambda_2) \\ 2 & w.p.~\lambda_2/(\lambda_1+\lambda_2)\end{array}\right.$, then $Y\sim\sum_{n=1}^N M_n$, a compound poisson distribution.
To prove the result, note that the number of times -3 is added to $Y$ in $Y = -3X_1 + 2X_2$ is poisson distributed with rate $\lambda_1$, and the number of times 2 is added is poisson distributed with rate $\lambda_2$, and the counts are independent. By the properties of merged poisson processes, this is exactly the case with our merged poisson process.