= Solution
Let $\mathbf a=d\mathbf u/dt$ denote coordinate <acceleration> and use $A^\mu$ for <four-acceleration>, to distinguish these two objects. Applying the <chain rule> to the <Lorentz factor> gives
$$
\frac{d\gamma_u}{dt}
=-\frac12\left(1-\frac{\mathbf u\cdot\mathbf u}{c^2}\right)^{-3/2}
\left(-\frac{2\mathbf u\cdot\mathbf a}{c^2}\right)
=\boxed{\frac{\gamma_u^3\mathbf u\cdot\mathbf a}{c^2}}.
$$
Since $dt/d\tau=\gamma_u$, differentiation of the <four-velocity> $U^\mu=\gamma_u(c,\mathbf u)$ with respect to <proper time> yields
$$
A^\mu=\gamma_u\frac{d}{dt}\big[\gamma_u(c,\mathbf u)\big]
=\boxed{\left(\frac{\gamma_u^4\mathbf u\cdot\mathbf a}{c},\
\gamma_u^2\mathbf a+\frac{\gamma_u^4(\mathbf u\cdot\mathbf a)\mathbf u}{c^2}\right)}.
$$
This is also $\gamma_u^2[\gamma_u^2(\mathbf u\cdot\mathbf a)/c,\ \mathbf a+\gamma_u^2(\mathbf u\cdot\mathbf a)\mathbf u/c^2]$, as required.
The invariant <Minkowski norm> of the <four-velocity> is $U^\mu U_\mu=c^2$. Differentiating this identity gives $U^\mu A_\mu=0$. In the instantaneous <rest frame>, $U^\mu=(c,\mathbf0)$, so $A^0=0$ and the spatial components equal the physically experienced <proper acceleration> $\boldsymbol\alpha$. Therefore $\alpha^2=-A^\mu A_\mu$ in the $+---$ <metric signature>.
For a direct calculation, put $q=\mathbf u\cdot\mathbf a$ and expand the temporal and spatial squares:
$$
\begin{aligned}
\alpha^2
&=\left|\gamma_u^2\mathbf a+\frac{\gamma_u^4q\mathbf u}{c^2}\right|^2
-\frac{\gamma_u^8q^2}{c^2}\\
&=\gamma_u^4a^2+\frac{2\gamma_u^6q^2}{c^2}
+\frac{\gamma_u^8q^2}{c^2}\left(\frac{u^2}{c^2}-1\right)\\
&=\boxed{\gamma_u^4\mathbf a\cdot\mathbf a+\frac{\gamma_u^6(\mathbf u\cdot\mathbf a)^2}{c^2}}.
\end{aligned}
$$
The last step uses $1-u^2/c^2=\gamma_u^{-2}$. This <proper acceleration from coordinate velocity and acceleration> also reads $\alpha^2=\gamma_u^6a_\parallel^2+\gamma_u^4a_\perp^2$, providing the familiar one-dimensional limit $\alpha=\gamma_u^3|a_\parallel|$. Each expression has units $\mathrm{m^2\,s^{-4}}$.
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