From the system of differential equations
$$
\pmatrix{g&f\\-fg&1+f^2\\-f&g\\1+g^2&-fg}
\pmatrix{f''\\g''}
=
\pmatrix{6f'g'\\-3gf'^2+3ff'g'\\−3f'^2+3g'^2\\3gf'g'−3fg'^2}
$$


the first and third equations can be combined to
$$
\pmatrix{f&-g\\g&f}\pmatrix{f''\\g''}=3\pmatrix{f'^2-g'^2\\2f'g'}
\\\text{or}\\
(f+ig)(f''+ig'')=3(f'+ig')^2
$$
which gives by integrating once $$f'+ig'=C(f+ig)^3\tag{\*},$$ and twice
$$
(f+ig)^{-2}=D-Cu\implies f^2+g^2=|D-Cu|^{-1}.
$$


Multiplying 4th,resp.2nd equation with $f,$ resp. $g$ and summing gives
$$
ff''+gg''=-3(f^2g'^2-2ff'gg'+f'^2g^2)=-3(fg'-f'g)^2=-3Im((f-ig)(f'+ig'))^2
$$
But also 
$$
ff''+gg''=Re((f-ig)(f''+ig''))=3\frac{Re\Bigl((f-ig)(f'+ig'))^2\Bigr)}{f^2+g^2}
$$
If we insert the above first-order differential equation (\*) we get to the identity
$$
-\operatorname{Im}(C(f^2+g^2)(f+ig)^2)^2=\frac{\operatorname{Re}\Bigl((C(f^2+g^2)(f+ig)^2)^2\Bigr)}{f^2+g^2}
\\\iff\\
-(f^2+g^2)\operatorname{Im}(C(f+ig)^2)^2=\operatorname{Re}(C^2(f+ig)^4)=Re(C(f+ig)^2)^2-Im(C(f+ig)^2)^2
\\\iff\\
(1-|Cu-D|^{-1})\operatorname{Im}(C/(Cu-D))^2=\operatorname{Re}(C/(Cu-D))^2
$$
Solution says that the last should be impossible for $C\neq 0$. Can someone explain me why is this impossible? It is clear if $C$ is real, but I believe it can be complex. Functions $f$ and $g$ are real functions of the same, real parameter.



Why is $(1-|Cu-D|^{-1})\operatorname{Im}(C/(Cu-D))^2=\operatorname{Re}(C/(Cu-D))^2$ impossible for $C \neq 0$
Let us assume that $C\neq 0$. Multiplying the last display by $\left|u-\frac{D}{C}\right|^4$, we arrive at
$$\left(1-\frac{1}{|Cu-D|}\right)\left(\operatorname{Im}\frac{D}{C}\right)^2=\left(u-\operatorname{Re}\frac{D}{C}\right)^2.$$
We claim that this equation has at most $6$ solutions $u\in\mathbb{R}$. Indeed, we have that
$$\frac{-1}{|Cu-D|}\left(\operatorname{Im}\frac{D}{C}\right)^2=\left(u-\operatorname{Re}\frac{D}{C}\right)^2-\left(\operatorname{Im}\frac{D}{C}\right)^2,$$
$$\frac{1}{(Cu-D)(\bar C u-\bar D)}\left(\operatorname{Im}\frac{D}{C}\right)^4=\left(\left(u-\operatorname{Re}\frac{D}{C}\right)^2-\left(\operatorname{Im}\frac{D}{C}\right)^2\right)^2,$$
$$\left(\operatorname{Im}\frac{D}{C}\right)^4=(Cu-D)(\bar C u-\bar D)\left(\left(u-\operatorname{Re}\frac{D}{C}\right)^2-\left(\operatorname{Im}\frac{D}{C}\right)^2\right)^2.$$
The right hand side is a real polynomial in $u$ of degree $6$, which proves the claim.


