Finding a holomorphic function with real part $u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$. I am asked to 

find the holomorphic function $f(z)=f(x\pm iy)$ with real part
  $$u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$$
  and such that $f(0)=0$.

I have tried to apply Cauchy-Riemann equations directly but I don't seem to be getting anywhere. I've also tried writing $u$ in terms of $z$ and $\bar{z}$ and then applying $f'=2\partial_z{u}$, but the expression of the derivative seems unmanagable. Since $z$ and $\bar{z}$ are symmetric in the expression of $u$, I also have the relation $\partial_z{u}=\overline{\partial_{\overline{z}}u}$, which I've tried to plug into $\partial_{\overline{z}}f=0$. But I dont know how to proceed from here.

Hint: In general, use these substitutions
\begin{align}
x=\frac{z+\bar{z}}{2}\\
y=\frac{z-\bar{z}}{2i}\\
|z|^2=x^2+y^2\\
=z\bar{z}
\end{align}
Edit: After substitution
\begin{align}
u
&= \dfrac{\frac{z+\bar{z}}{2}\left(1+z\overline{z}\right)}{1+z^2+\overline{z}^2+z^2\overline{z}^2}\\
&= \dfrac12\dfrac{z(1+\overline{z}^2)+\overline{z}(1+z^2)}{(1+z^2)(1+\overline{z}^2)}\\
&= \dfrac12\left(\dfrac{z}{1+z^2}+\dfrac{\overline{z}}{1+\overline{z}^2}\right)
\end{align}
then
$$v=\dfrac{1}{2i}\left(\dfrac{z}{1+z^2}-\dfrac{\overline{z}}{1+\overline{z}^2}\right)$$
and
$$\color{blue}{f(z)=\dfrac{z}{1+z^2}}$$
