= Solution
The two <Euler-Lagrange equations> are
$$
\frac d{dx}\frac{\partial f}{\partial u'}-\frac{\partial f}{\partial u}=0,
\qquad
\frac d{dx}\frac{\partial f}{\partial w'}-\frac{\partial f}{\partial w}=0.
$$
Differentiating $\kappa=f-u'f_{u'}-w'f_{w'}$ along an extremal and using these equations gives $d\kappa/dx=f_x$. Hence the <Beltrami identity> makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a <cyclic coordinate> and the corresponding momentum $f_{u'}$ or $f_{w'}$ is an additional <first integral>; continuous symmetries give the analogous conserved quantities through <Noether's theorem>.
Put $A=1-m/u$. Since $w$ is cyclic,
$$
2Aw'=2\lambda,
\qquad Aw'=\lambda
$$
for a constant $\lambda$. The Beltrami integral is
$$
\kappa=f-u'f_{u'}-w'f_{w'}
=A^{-1}u'^2-Aw'^2
=A^{-1}(u'^2-\lambda^2).
$$
Therefore
$$
\boxed{u'^2=\lambda^2+\kappa\left(1-\frac mu\right)}.
$$
If $\kappa=-\lambda^2$, then $u'^2=\lambda^2m/u$ and
$$
\frac{dw}{du}=\pm\frac{(u/m)^{3/2}}{u/m-1}.
$$
For $z>1$, define
$$
F(z)=\frac23z^{3/2}+2z^{1/2}
+\log\frac{\sqrt z-1}{\sqrt z+1};
$$
the hinted identity gives $F'(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to+\infty$ as $z\to\infty$, the prescribed limit selects the minus sign:
$$
\boxed{w(u)=C-m\left[
\frac23\left(\frac um\right)^{3/2}
+2\sqrt{\frac um}
+\log\frac{\sqrt{u/m}-1}{\sqrt{u/m}+1}
\right]}.
$$
As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to+\infty}$.
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