Im(z) and Re(z) of $\bar z^2+ \frac {1}{z^2}$
Solution 1:
Let's assume $z=re^{i\varphi}$. We have then $$ \bar{z}^2 + \frac{1}{z^2} = e^{-2i\varphi} (r^2+ \frac{1}{r^2})$$ Knwing $\bar{z}^2 + \frac{1}{z^2}$ you can find its module and argument, and from them you can calculate $r$ and $\varphi$ respectively. Then you have ${\rm Im}(z) = r\sin\varphi$ and ${\rm Re}(z) = r\cos\varphi$.
Solution 2:
You are in the right track for finding the Im and Re of $\bar z^2+ \frac {1}{z^2}$. Just note that $z\bar z=x^2+y^2\in \mathbb R$ and this simplifies your calculations hugely.