[proofplan]
Set $M:=L(\lambda)\otimes L(\mu)$ and interpret the desired multiplicity inside the finite-dimensional $\mathfrak g$-module $M$. Complete reducibility reduces the multiplicity computation to the dimension of $\operatorname{Hom}_{\mathfrak g}(L(\nu),M)$, using [Schur's lemma](/theorems/2414) on irreducible summands. The Hom-tensor adjunction then identifies this Hom space with the $\mathfrak g$-invariant subspace of $L(\nu)^*\otimes M$, which is exactly the stated [vector space](/page/Vector%20Space).
[/proofplan]
[step:Reduce the multiplicity to a Hom space]
Define the finite-dimensional $\mathfrak g$-module
\begin{align*}
M:=L(\lambda)\otimes L(\mu),
\end{align*}
where $\mathfrak g$ acts by the [tensor product](/page/Tensor%20Product) action.
By [Weyl's complete reducibility theorem](/theorems/3755) for finite-dimensional representations of complex semisimple Lie algebras (citing a result not yet in the wiki: Complete reducibility of finite-dimensional representations of complex semisimple Lie algebras), the finite-dimensional $\mathfrak g$-module $M$ is semisimple. Hence there exists a finite-dimensional $\mathfrak g$-module $N$ and an integer $c_{\lambda\mu}^{\nu}\in\mathbb Z_{\ge 0}$ such that
\begin{align*}
M\cong L(\nu)^{\oplus c_{\lambda\mu}^{\nu}}\oplus N
\end{align*}
as $\mathfrak g$-modules, where no irreducible direct summand of $N$ is isomorphic to $L(\nu)$.
Applying the complex vector space functor $\operatorname{Hom}_{\mathfrak g}(L(\nu),-)$ gives
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),M)
\cong
\operatorname{Hom}_{\mathfrak g}(L(\nu),L(\nu)^{\oplus c_{\lambda\mu}^{\nu}})
\oplus
\operatorname{Hom}_{\mathfrak g}(L(\nu),N).
\end{align*}
Since $L(\nu)$ is irreducible, Schur's lemma over $\mathbb C$ gives
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_{\mathfrak g}(L(\nu),L(\nu))=1
\end{align*}
and, for every irreducible summand $S$ of $N$,
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),S)=0.
\end{align*}
Thus
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_{\mathfrak g}(L(\nu),M)=c_{\lambda\mu}^{\nu}.
\end{align*}
[guided]
The multiplicity $c_{\lambda\mu}^{\nu}$ means the number of copies of $L(\nu)$ appearing in a semisimple decomposition of $M:=L(\lambda)\otimes L(\mu)$. The module $M$ is finite-dimensional because both $L(\lambda)$ and $L(\mu)$ are finite-dimensional. Since $\mathfrak g$ is complex semisimple, Weyl's complete reducibility theorem for finite-dimensional $\mathfrak g$-modules applies (citing a result not yet in the wiki: Complete reducibility of finite-dimensional representations of complex semisimple Lie algebras). Therefore $M$ decomposes as a finite [direct sum](/page/Direct%20Sum) of irreducible $\mathfrak g$-modules.
We isolate the summands isomorphic to $L(\nu)$. This gives a decomposition
\begin{align*}
M\cong L(\nu)^{\oplus c_{\lambda\mu}^{\nu}}\oplus N,
\end{align*}
where $N$ is a semisimple finite-dimensional $\mathfrak g$-module with no irreducible direct summand isomorphic to $L(\nu)$. The point of this decomposition is that maps out of an irreducible module detect exactly the copies of that irreducible module.
Apply $\operatorname{Hom}_{\mathfrak g}(L(\nu),-)$ to the direct sum. Since Hom into a finite direct sum is the corresponding direct sum of Hom spaces,
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),M)
\cong
\operatorname{Hom}_{\mathfrak g}(L(\nu),L(\nu)^{\oplus c_{\lambda\mu}^{\nu}})
\oplus
\operatorname{Hom}_{\mathfrak g}(L(\nu),N).
\end{align*}
By Schur's lemma over $\mathbb C$, every $\mathfrak g$-endomorphism of the irreducible module $L(\nu)$ is scalar, so
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_{\mathfrak g}(L(\nu),L(\nu))=1.
\end{align*}
The same lemma also says that if $S$ is irreducible and not isomorphic to $L(\nu)$, then
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),S)=0.
\end{align*}
Since every irreducible summand of $N$ is non-isomorphic to $L(\nu)$, it follows that
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),N)=0.
\end{align*}
Therefore the Hom space has one complex dimension for each copy of $L(\nu)$ in $M$, and no contribution from the remaining summands:
\begin{align*}
\dim_{\mathbb C}\operatorname{Hom}_{\mathfrak g}(L(\nu),M)=c_{\lambda\mu}^{\nu}.
\end{align*}
[/guided]
[/step]
[step:Identify the Hom space with invariant tensors]
The modules $L(\nu)$ and $M$ are finite-dimensional $\mathfrak g$-modules. By [citetheorem:9392], applied with $V=L(\nu)$ and $W=M$, there is a natural complex vector space isomorphism
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),M)
\cong
\left(L(\nu)^*\otimes M\right)^{\mathfrak g}.
\end{align*}
Here $L(\nu)^*$ carries the contragredient action, and for any $\mathfrak g$-module $E$ the invariant subspace is
\begin{align*}
E^{\mathfrak g}:=\{e\in E:x\cdot e=0\text{ for every }x\in\mathfrak g\}.
\end{align*}
Substituting $M=L(\lambda)\otimes L(\mu)$ gives
\begin{align*}
\operatorname{Hom}_{\mathfrak g}(L(\nu),L(\lambda)\otimes L(\mu))
\cong
\left(L(\nu)^*\otimes L(\lambda)\otimes L(\mu)\right)^{\mathfrak g}.
\end{align*}
[/step]
[step:Take dimensions to obtain the multiplicity formula]
Combining the dimension identity from the first step with the isomorphism from the second step yields
\begin{align*}
c_{\lambda\mu}^{\nu}
=
\dim_{\mathbb C}\operatorname{Hom}_{\mathfrak g}(L(\nu),M)
=
\dim_{\mathbb C}\left(L(\nu)^*\otimes L(\lambda)\otimes L(\mu)\right)^{\mathfrak g}.
\end{align*}
This is the asserted formula for the multiplicity of $L(\nu)$ in $L(\lambda)\otimes L(\mu)$.
[/step]
