= Solution
In the defining representation of the <symplectic Lie algebra> $\mathfrak{sp}_4$, the weights are $\varepsilon_1,\varepsilon_2,-\varepsilon_1,-\varepsilon_2$. Their <highest weight>, for the <root basis> of part (b), is therefore
$$
\boxed{\operatorname{highest\ weight}(V)=\varepsilon_1=\omega_1.}
$$
The flip on the <tensor square> commutes with the action of $\mathfrak{sp}_4$, giving $V\otimes V=\operatorname{Sym}^2V\oplus\Lambda^2V$ with dimensions $10$ and $6$.
If $v_1$ is a <highest-weight vector> of weight $\varepsilon_1$, then $v_1\otimes v_1$ is a <highest-weight vector> of weight $2\omega_1$ in the <symmetric square>. By <Weyl complete reducibility theorem>, this ensures an irreducible summand $V(2\omega_1)$ occurs. Its dimension is $10$ by part (b), exhausting the <symmetric square>.
Let $\omega$ denote the preserved <symplectic form>. The <symplectic contraction of an exterior square> is the nonzero equivariant map
$$
c:\Lambda^2V\longrightarrow\mathbb C,\qquad c(v\wedge w)=\omega(v,w),
$$
where the target is a <trivial Lie algebra representation>. Thus its <kernel> has dimension $5$. Choose a weight vector $v_2$ of weight $\varepsilon_2$ in a <symplectic basis>, with $\omega(v_1,v_2)=0$. The vector $v_1\wedge v_2$ belongs to this <kernel> and has weight $\varepsilon_1+\varepsilon_2=\omega_2$. It is a <highest-weight vector>: none of $\omega_2+\alpha$, for $\alpha\in\Phi^+$, is a weight of $\Lambda^2V$, whose weights are $\pm\varepsilon_1\pm\varepsilon_2$ and zero. Hence the <kernel> contains $V(\omega_2)$; its dimension $5$ exhausts the <kernel>. <Weyl complete reducibility theorem> supplies a complementary invariant line $V(0)$.
The resulting <tensor-square decomposition of the defining sp4 representation> is
$$
\boxed{V\otimes V\cong V(2\omega_1)\oplus V(\omega_2)\oplus V(0),\qquad16=10+5+1.}
$$
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