One way to know which are the simple modules appearing in the decomposition of $M=\mathfrak{sl}_3$ as a $\mathfrak{g}=\mathfrak{sl}_2$ module is to look at the eigenvalues of the action of $H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ on $M$.
An easy computation, which I hope I did correctly, shows that they are $2,\quad 1,\quad1,\quad0,\quad0,\quad-1,\quad-1,\quad-2.$ We can collect this information in a polynomial in the variable $q$, summing $q^\lambda$ for all eigenvalues $\lambda$, to obtained the character $\chi_M$ of $M$: in this case we get $\chi_M=q^2+2q+2+2q^{-1}+q^{-2}.$
Now, the character is additive with respect to direct sums, and in fact uniquely identifies the isomorphism class of modules. Moreover, for each $\ell\geq1$, the character of the unique simple $\mathfrak{g}$-module $V_\ell$ of dimension $\ell$ is $\chi_{V_\ell}=\frac{q^\ell-q^{-\ell}}{q-q^{-1}}.$
Now, an easy computation shows that $\chi_M=\chi_{V_3}+2\chi_{V_2}+\chi_{V_1}.$ The above remarks imply that $M\cong V_3\oplus V_2\oplus V_2\oplus V_1.$
Now of course you may want to do this explicitly, actually finding the submodules---people want all sort of wierd things! In this case it is not difficult to find them by looking.
The $\mathfrak{g}$-module $V_3$, the unique simple module of dimension $3$, is of course isomorphic to $\mathfrak g$ with its adjoint action. It is immediate that inside out module $M$ there is a copy of $\mathfrak{g}$.
Next, the subspace of matrices of the form $\begin{pmatrix}0&0&*\\0&0&*\\0&0&0\end{pmatrix}$ is easily seen to be a $\mathfrak{g}$-module isomorphic to $V_2$ (which is the "taulotogical" module), just as the subspace of matrices of the form $\begin{pmatrix}0&0&0\\0&0&0\\*&*&0\end{pmatrix}$.
Finally, the matrix $\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}$ spans a subspace of matrices which commute with $\mathfrak{g}$.