= Solution
First use the additional admitted form of the <efficient score>. Write $\widetilde\ell=e\zeta(X)$ and let $\tau^2=E_f\varepsilon^2>0$. The difference $\dot\ell-\widetilde\ell$ is the <orthogonal projection> onto the <nuisance tangent space>. Part (c) makes every $e\psi(X)$ orthogonal to that space. Therefore, for all $\psi\in L^2(v)$,
$$
0=E[(h_\theta(X)\rho(\varepsilon)-\varepsilon\zeta(X))\varepsilon\psi(X)]
=E_v\bigl[\{h_\theta E_f(\varepsilon\rho)-\tau^2\zeta\}\psi\bigr].
$$
The <function> in braces belongs to <L2 space>; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction is
$$
\boxed{\widetilde\ell_{\theta,\eta}(x,e)
=-e\,\frac{\int s f'(s)\,ds}{\int s^2f(s)\,ds}\,h_\theta(x).}
$$
Under the regular tail condition $s f(s)\to0$ at both infinities, <integration by parts> gives $\int s f'(s)\,ds=-1$, and the formula becomes $e h_\theta(x)/\tau^2$. This also confirms the sign.
\b[The admitted product form is an extra restriction; it does not follow for every independent-error regression model.] To locate the restriction precisely, suppose the error-density nuisance <statistical paths> preserve both normalization and mean to first order. Their closed <mean-preserving error tangent space> is $\{\gamma:E_f\gamma=E_f(\varepsilon\gamma)=0\}$. The full <nuisance tangent space> is the orthogonal sum of this space and the centered <functions> of $X$.
For completeness, bounded <functions> satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its <expected value>, and then subtract a multiple of a fixed bounded centered <function> $q$ with $E_f(\varepsilon q)\ne0$. Such a $q$ exists by truncating $\varepsilon$, since $\tau^2>0$. The correction coefficients tend to zero by the <Cauchy-Schwarz inequality>, so the corrected truncations converge in <L2 space>.
Let $m=E_vh_\theta(X)$. With $E_f\rho=0$ and $E_f(\varepsilon\rho)=1$, the <orthogonal projection> of $h_\theta(X)\rho(\varepsilon)$ onto the error nuisance space is $m\{\rho(\varepsilon)-\varepsilon/\tau^2\}$; its projection onto the covariate nuisance space is zero. Thus the general <efficient score in independent-error regression> is
$$
\boxed{\widetilde\ell=(h_\theta(X)-m)\rho(\varepsilon)+\frac{m\varepsilon}{\tau^2},
\qquad
\widetilde I=\operatorname{Var}_v(h_\theta)E_f\rho^2+\frac{m^2}{\tau^2}.}
$$
The <orthogonality> of the two terms gives the displayed <efficient information>. For a <normal distribution> of the error, $\rho(e)=e/\tau^2$, so this reduces to the admitted formula. It also does so when $h_\theta$ is constant.
A concrete counterexample to the generality of the admission is $g_\theta(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard <logistic distribution> error. Here $m=0$ and $\rho(e)=\tanh(e/2)$, so the actual <efficient score> is $x\tanh(e/2)$. It cannot equal $e\zeta(x)$ because $\tanh(e/2)/e$ is not constant. This <score function> is already orthogonal to every nuisance <score function>: its factor $X$ is centered against error-only directions, and its factor $\rho(\varepsilon)$ is centered against covariate-only directions. \b[The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.]
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