= Solution
Here $[M]_t=t$ and $dA_t=dt$, so the formula in part (b) gives
$$
A_t^f=\int_0^t g(B_s,s)\,ds,\qquad g(x,s)=f_y(x,s)+\tfrac12f_{xx}(x,s).
$$
If $g$ vanishes on $\mathbb R\times[0,\infty)$, this integral is zero. Conversely suppose $A_t^f=0$ for all $t$ almost surely. Its integrand is continuous along each path; the <fundamental theorem of calculus> implies $g(B_t,t)=0$ for all $t$ on that event. For any fixed $t>0$, the <normal distribution> of $B_t$ has a strictly positive density at every real point. Since $g(\cdot,t)$ is continuous, a nonzero value at any spatial point would give a nonzero value on an open interval of positive Brownian probability, a contradiction. Thus $g(x,t)=0$ for every $x$ and every positive $t$. Continuity extends this to $t=0$.
Consequently the exact condition is
$$
\boxed{f_y(x,y)+\frac12f_{xx}(x,y)=0\quad\text{for every }x\in\mathbb R,\ y\geq0.}
$$
These are <space-time harmonic functions along Brownian motion>, satisfying the <backward heat equation>. The second coordinate never takes negative values, so no equation is imposed for $y<0$, beyond compatibility with the assumed global $C^2$ regularity. The normalized finite-variation term is unique by part (a), so it cannot be removed by choosing a different decomposition.
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