= Solution
A scalar <M-estimator> is a selected root of $n^{-1}\sum_i\psi(Y_i,\widehat\theta)=0$, or equivalently a minimizer of a differentiable empirical objective whose derivative is this estimating equation. Its population <statistical functional> $T(F)=\theta_F$ solves $\int\psi(x,\theta_F)dF(x)=0$. Assume local uniqueness, consistency of the sample root, a valid differentiable local expansion, finite $\mathbb E_F\psi^2$, and
$$
A=\int\partial_\theta\psi(x,\theta_F)dF(x)\ne0.
$$
These conditions include domination sufficient to differentiate under the integral and the required local uniform control of the sample derivative.
For $F_\varepsilon=(1-\varepsilon)F+\varepsilon\delta_z$, differentiate the population estimating equation at $\varepsilon=0$. Since its uncontaminated integral is zero,
$$
0=\psi(z,\theta_F)+A\left.\frac{dT(F_\varepsilon)}{d\varepsilon}\right|_{0}.
$$
Thus the <influence function> is \b[$\operatorname{IF}(z;T,F)=-\psi(z,\theta_F)/A$], with a proportionality constant <independent> of $z$.
Expanding the sample equation at its population root gives
$$
\sqrt n(\widehat\theta-\theta_F)=-\frac1A\frac1{\sqrt n}\sum_i\psi(Y_i,\theta_F)+o_p(1).
$$
The summands have mean zero. The <central limit theorem> and <Slutsky's theorem> therefore yield the scalar <sandwich variance of an M-estimator>
$$
\boxed{\sqrt n(\widehat\theta-\theta_F)\Rightarrow N(0,V(\psi,F)),\qquad V(\psi,F)=\frac{\int\psi(x,\theta_F)^2dF(x)}{[\int\partial_\theta\psi(x,\theta_F)dF(x)]^2}.}
$$
This $V$ is the limiting <variance> of the $\sqrt n$-scaled error; the estimator's <variance> is approximately $V/n$. In a correctly centred <location family> with $\psi(x,\theta)=\psi(x-\theta)$, let $m=\mathbb E_F\psi'(X-\theta)$. Then $A=-m$, so the <influence function> is $\psi(z-\theta)/m$ and $V=\mathbb E\psi^2/m^2$. The minus sign from differentiating $x-\theta$ is essential.
The clipped <Huber score> in the subsequent subparts requires \b[$0<b<\infty$]. The PDF only displays an upper bound: at $b=0$ the estimating function vanishes identically, and at $b<0$ it is a constant with no root. A positive clipping threshold is the regular interpretation.
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