= Solution
Now $p_{\theta,\eta}(x,y)=v(x)f(y-g_\theta(x))$. With $h_\theta=\partial_\theta g_\theta$ and the <location score> $\rho(e)=-f'(e)/f(e)$, differentiation of the <log-likelihood> gives
$$
\boxed{\dot\ell_{\theta,\eta}(x,y)=-h_\theta(x)\frac{f'(e)}{f(e)}=h_\theta(x)\rho(e).}
$$
Take the usual regularity conditions $h_\theta\in L^2(v)$ and $\rho\in L^2(f)$ so that this <score function> belongs to <L2 space>.
For the specified nuisance <statistical path>, differentiating $\log(1+t\gamma(e))$ at zero gives the <score function> $\gamma(e)$. Normalization requires $E_f\gamma=0$. There is a second constraint: the <statistical path> must preserve the model's zero error <expected value>. Thus
$$
0=\int e f(e)(1+t\gamma(e))\,de=t\int e\gamma(e)f(e)\,de,
\qquad\boxed{E_f[\varepsilon\gamma(\varepsilon)]=0.}
$$
This constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric <logistic distribution>, for example, $\gamma(e)=\tanh(e)$ is bounded and has zero <expected value>, but $E[\varepsilon\tanh(\varepsilon)]>0$; it would not preserve the mean.
For any $\psi\in L^2(v)$, <independence> implies
$$
E[\gamma(\varepsilon)\varepsilon\psi(X)]
=E_f[\varepsilon\gamma(\varepsilon)]E_v\psi(X)=\boxed{0}.
$$
All these products are integrable by the <Cauchy-Schwarz inequality>, since $\varepsilon\psi(X)$ is <square-integrable>. Thus every such error-density <score function> is orthogonal to every $e\psi(x)$. Covariate-density nuisance <score functions> $b(X)$ are also orthogonal to these <functions>, since $E\varepsilon=0$. This is the <mean-preserving error tangent space> orthogonality used in the final calculation.
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