Theorem.  Every solution of the equation (*) uu" &minus u'<sup>2</sup> = e<sup>z</sup> (1 &minus u<sup>4</sup>) is meromorphic in the entire plane.
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Step one is to replace the equation with a system of three equations that eliminates the division by u, making it easier to handle zeroes and poles.  Let v represent u' and w let represent (u" &minus u')/u.  Then the first two equations are obvious: u' = v and v' = uw + v.  If we rewrite the original equation as (**)
e<sup>-z</sup>(u<sup>2</sup>w + uv &minus v<sup>2</sup>) = 1 &minus u<sup>4</sup>, differentiation and simplification give the third: w' = &minus 4e<sup>z</sup>uv.  Note that if u, v, and w satisfy these three equations, then u may satisfy the more general equation uu" &minus u'<sup>2</sup> = e<sup>z</sup> (a &minus u<sup>4</sup>) for a constant a other than 1.  But if a = 1 at one point, they are equal everywhere.
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Step two is to note that if we let u<sub>1</sub> = 1/u, v<sub>1</sub> = u<sub>1</sub>' = &minus v/u<sup>2</sup> and w<sub>1</sub> = (u<sub>1</sub>" &minus u<sub>1</sub>')u<sub>1</sub> = 2(u'/u)<sup>2</sup> &minus w, then u<sub>1</sub>, v<sub>1</sub>, and w<sub>1</sub> satisfy exactly the same equations (including the original (*)).
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According to the standard existence theorem, 2.2.2 of Hille's Ordinary Differential Equations in the Complex Domain, the system has a unique solution through each point (z<sub>0</sub>, u<sub>0</sub>, v<sub>0</sub>, w<sub>0</sub>).  Clearly every zero of u corresponds to a pole of u<sub>1</sub> and vice versa.
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Each solution extends uniquely to a complete Riemann surface in (z, u, v, w) space.  We can think of the pair ((z, u, v, w), (z, u<sub>1</sub>, v<sub>1</sub>, w<sub>1</sub>)) as a sphere-valued solution, with u = 0 representing the south pole and u<sub>1</sub> = 0 representing the north pole (or infinity of the extended plane).  This trick allows us to consider poles as regular points, and if we can show that there are no other singularities, then the projection to z would be a covering of the plane, and therefore 1-1, making the solution everywhere meromorphic.
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Since each solution is a complete submanifold, other singularities correspond to places where v and/or w become infinite.  However, if we pick an r > 1 and restrict to a closed region where |u| &le r and let k be a bound for 4e<sup>|z|</sup>|u|, then |v'| &le r|w| + |v| and |w'| &le k|v|.  Consider the equations u' = v, v' = rw + v, and w' = kv, then the solution through (|z<sub>0</sub>|, |u<sub>0</sub>|, |v<sub>0</sub>|, |w<sub>0</sub>|) will give bounds for |v| and |w|, showing that such singularities do not exist on such regions.  But by switching to the reciprocal solution when necessary, we can cover the plane with the interiors of these regions.