Finding a holomorphic complex function We have that $\varphi:R->R$ can be derived twice, we want to find a function f=u+iv which is holomorphic. $u(x,y)=\varphi ( \frac{x^2+y^2}{x} )$
Solution:
We calculate that:
$u_{x}=\frac{x^2-y^2}{x^2} \varphi' (\frac{x^2+y^2}{x})$ and $u_{xx}= \frac{2y^{2}}{x}\varphi' (\frac{x^2+y^2}{x})+(\frac{x^2-y^{2}}{x^2})^2\varphi'' (\frac{x^2+y^2}{x})$
$u_{y}=\frac{2y}{x}\varphi'(\frac{x^2+y^2}{x})$ and $u_{yy}=\frac{2}{x}\varphi'(\frac{x^2+y^2}{x})+\frac{4y^2}{x^2}\varphi''(\frac{x^2+y^2}{x})$
I have to use this information and:
$$u_{xx}+u_{yy}=0$$
but I end up having this massive equaion which I can't solve:
$$\frac{2(y^2+1)}{x}\varphi'(\frac{x^{2}+y^2}{x})+((\frac{x^2-y^2}{x^2})^2+\frac{4y^2}{x^2})\varphi''(\frac{x^{2}+y^2}{x})=0$$
So I either misscalculated something or there is a magic trick which I am missing. Can you help me find my mistake and fix it ?

Note first that your differential equation (which contains a small typo, because the derivative of $-y^{2}/x^{2}$ with respect to $x$ is $2y^{2}/x^{3}$) can be written
\begin{align*}
  0 &= \frac{2(y^{2} + x^{2})}{x^{3}} \varphi'\left(\frac{x^{2} + y^{2}}{x}\right)
  + \left[\left(\frac{x^{2}-y^{2}}{x^{2}}\right)^{2} + \frac{4y^{2}}{x^{2}}\right] \varphi''\left(\frac{x^{2}+y^{2}}{x}\right) \\
  &= \frac{2(y^{2} + x^{2})}{x^{3}} \varphi'\left(\frac{x^{2} + y^{2}}{x}\right)
  + \left(\frac{x^{2} + y^{2}}{x^{2}}\right)^{2} \varphi''\left(\frac{x^{2}+y^{2}}{x}\right).
  \tag{*}
\end{align*}
In particular, (*) holds where $y = 0$, and here it reduces to
$$
\frac{2}{x} \varphi'(x) + \varphi''(x) = 0.
$$
The solutions are easily checked to have the form
$$
\varphi(x) = c_{1} + \frac{c_{2}}{x},
$$
for which
$$
u(x, y) =  c_{1} + c_{2}\, \frac{x}{x^{2} + y^{2}}
$$
is the real part of $f(z) = c_{1} + \dfrac{c_{2}}{z}$.
