= Solution
The induced <raising operator> and <lowering operator> act by
$$
Ev_i=i\,v_{i-1},\qquad Fv_i=(4-i)v_{i+1},\qquad X(u\wedge v)=(Xu)\wedge v+u\wedge(Xv).
$$
Since $Ew_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible <sl2 Lie algebra> module $\operatorname{Sym}^6V$. By <complete reducibility of semisimple Lie algebra representations>, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the <weight diagram> leaves weights $2,0,-2$ once each. Hence
$$
\boxed{U\cong\operatorname{Sym}^6V\oplus\operatorname{Sym}^2V.}
$$
These are the only two irreducible isomorphism types occurring as submodules.
For the three-dimensional summand, calculate $Ew_{03}=3w_{02}$ and $Ew_{12}=w_{02}$. Therefore
$$
\boxed{v_+=w_{03}-3w_{12}}
$$
is a nonzero <highest-weight vector> of weight $2$. The <classification of finite-dimensional sl2 representations> shows that the submodule $W$ it generates is the irreducible module $\operatorname{Sym}^2V$, of dimension three.
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