finding inverse Z transform with usage of tables (LTI and causal sequences)
Your solution is correct. This is the way I see it (assuming that the sequence is causal/right-sided):

$$\mathcal{Z}^{-1}\left\{\frac{3}{z+2}\right\}[n]=3\mathcal{Z}^{-1}\left\{z^{-1}\frac{z}{z+2}\right\}[n]=\\=3\mathcal{Z}^{-1}\left\{\frac{z}{z+2}\right\}[n-1]=3(-2)^{n-1}u[n-1]$$