= Solution
Let
$$
v=B_{\tau^{-1}f}(z/\tau).
$$
The <proximal operator> optimality condition says
$$
z/\tau-v\in\tau^{-1}\partial f(v),
\quad\text{so}\quad p:=z-\tau v\in\partial f(v).
$$
By <subgradient inversion under convex conjugacy>, which uses <Fenchel–Young inequality> and $f^{**}=f$ from the <Fenchel-Moreau theorem>,
$$
v\in\partial f^*(p).
$$
Consequently $z-p=\tau v\in\tau\partial f^*(p)$, meaning $p=B_{\tau f^*}(z)$. Uniqueness of the <backward subgradient step> identifies the result:
$$
\boxed{B_{\tau f^*}(z)=z-\tau B_{\tau^{-1}f}(z/\tau).}
$$
This is the scaled <Moreau decomposition>. It requires \b[one backward step on $f$, with reciprocal parameter $1/\tau$ and scaled input $z/\tau$], followed by a scalar multiplication and subtraction. No separate proximal computation of the <convex conjugate> is needed.
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