= Solution
Put $b=\delta B_\parallel/B_0$ and $\mathbf b_\perp=\delta\mathbf B_\perp/B_0=\hat{\mathbf z}\times\nabla_\perp\Psi/v_A$. Define the two-dimensional <Poisson bracket> by
$$
\{f,g\}=\partial_xf\,\partial_yg-\partial_yf\,\partial_xg.
$$
Then, to the retained anisotropic order, the derivative along the perturbed <magnetic field> is
$$
\boxed{\nabla_\parallel\equiv\frac{\mathbf B}{B_0}\cdot\nabla=\partial_z+\mathbf b_\perp\cdot\nabla_\perp=\partial_z+\frac1{v_A}\{\Psi,\cdot\}.}
$$
The omitted $b\partial_z$ term is smaller by $\epsilon$ than either retained term. The nonlinear field-line-bending term is of the same order as $\partial_z$, which is why it must remain.
Use $\mathbf u_e=-\alpha\nabla\times\mathbf B$ with $\alpha B_0=v_Ad_i$. The leading parallel and perpendicular <electron> <velocities> are
$$
\boxed{\mathbf u_{e\perp}=v_Ad_i\hat{\mathbf z}\times\nabla_\perp b,\qquad u_{e\parallel}=-d_i\nabla_\perp^2\Psi.}
$$
In detail, $\nabla\times(B_0b\hat{\mathbf z})=-B_0\hat{\mathbf z}\times\nabla_\perp b$, while the parallel <curl> of $B_0\hat{\mathbf z}\times\nabla_\perp\Psi/v_A$ is $B_0\nabla_\perp^2\Psi/v_A$. The perpendicular <curl> involving $\partial_z\delta\mathbf B_\perp$ is order-smaller. Exact <electron> <incompressibility> follows from the <divergence> of a <curl>; its small <velocity> correction supplies any <divergence> omitted by this leading representation.
The <electron magnetohydrodynamics> equation is frozen-in induction,
$$
\partial_t\mathbf B=(\mathbf B\cdot\nabla)\mathbf u_e-(\mathbf u_e\cdot\nabla)\mathbf B,
$$
because both $\mathbf B$ and $\mathbf u_e$ are <solenoidal>. Keeping its leading order-$\epsilon^2$ perpendicular terms gives
$$
\partial_t\mathbf b_\perp=\partial_z\mathbf u_{e\perp}+(\mathbf b_\perp\cdot\nabla_\perp)\mathbf u_{e\perp}-(\mathbf u_{e\perp}\cdot\nabla_\perp)\mathbf b_\perp.
$$
Let $X_f=\hat{\mathbf z}\times\nabla_\perp f=(-f_y,f_x)$. Expanding the two components gives $(X_f\cdot\nabla)X_g-(X_g\cdot\nabla)X_f=X_{\{f,g\}}$. Substitution therefore reduces the perpendicular equation to
$$
\frac1{v_A}X_{\partial_t\Psi}=v_Ad_iX_{\partial_zb}+d_iX_{\{\Psi,b\}}=v_Ad_iX_{\nabla_\parallel b}.
$$
The perpendicular-constant freedom in $\Psi$ has no magnetic effect and can be fixed so that
$$
\boxed{\partial_t\Psi=v_A^2d_i\nabla_\parallel b.}
$$
The parallel induction component at the same order is
$$
\partial_tb+\mathbf u_{e\perp}\cdot\nabla_\perp b=\nabla_\parallel u_{e\parallel}.
$$
The advection term vanishes since $\mathbf u_{e\perp}\cdot\nabla_\perp b=v_Ad_i\{b,b\}=0$. Thus
$$
\boxed{\partial_tb=-d_i\nabla_\parallel\nabla_\perp^2\Psi.}
$$
Together these are both required <reduced electron magnetohydrodynamics> equations. Terms such as $u_{e\parallel}\partial_zb$ and $b\partial_z\mathbf u_e$ are order-$\epsilon^3$ and consistently absent.
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