Let $\varphi_n$ denote the standart irreducible representation of $SU(2)$ group with highest weight $n$.
I know that irreducible representations of $\varphi_2 \otimes \varphi_3 = \varphi_5 \oplus \varphi_3 \oplus \varphi_1$ (according to Clebsh-Gordan decomposition).
What will be the invariant subspace of $\varphi_2 \otimes \varphi_3$?
What will be the explicit form of the actions of irreducible reps on invariant subspaces?
Working under the premise that you want to find, e.g. the submodule isomorphic to $\varphi_3$ inside the tensor product $\varphi_2\otimes\varphi_3$. Let the vectors ${\cal B}_2=\{x_0,x_1,x_2\}$ be a basis consisting of weight vectors of $\varphi_2$ (of respective weights 2,0,-2). Let similarly ${\cal B}_3=\{y_0,y_1,y_2,y_3\}$ be a basis of weight vectors of $\varphi_3$ (of respective weights 3,1,-1,-3). Then the space of vectors of weight three in the tensor product consists of the linear combinations $$ a x_0\otimes y_1+bx_1\otimes y_0, $$ where $a,b$ are arbitrary scalars. Calculate the effect of the raising operator (= a ladder operator that increases the weight) to such a linear combination. I dare not do that myself, because the details depend on how you normalized things and produced the bases ${\cal B}_2$ and ${\cal B}_3$ in the first place. You will find that - up to a scalar multiple - there is exactly one such linear combination that is annihilated by the ladder operator. That linear combination is then a high weight vector of weight three, and generates the copy $\varphi_3$.
You can similarly find (up to scalar multiple) a unique linear combination $$ cx_0\otimes y_2+dx_1\otimes y_1+ex_2\otimes y_0 $$ of weight one vectors that is annihilated by the raising operator. Tha generates the remaining summand $\varphi_1$ inside $\varphi_2\otimes\varphi_2$.