With two vector fields in cylindrical coordinates, I am trying to find how they may be perpendicular to each other
$$ A (\rho, \phi,z) = \rho cos \phi \hat\rho + \rho sin \phi \hat\phi + \rho \hat z $$ $$ B (\rho, \phi,z) = \rho cos \phi \hat\rho + \rho sin \phi \hat\phi - \rho \hat z$$
Any help or hints would be much appreciated
I think you mean $\hat{x}$ and $\hat{y}$ instead of $\hat{\rho}$ and $\hat{\phi}$, since $\rho \cos(\phi) = x$ and $\rho\sin(\phi) = y$ are the transformations to rectangular coordinates. I'm going to assume that what you meant is that $A = \rho_A \hat{\rho} + \phi_A \hat{\phi} + z_A \hat{z}$, and similarly for $B$, so that after changing coordinates, you have $A = \rho_A \cos(\phi_A) \hat{x} + \rho_A\sin(\phi_A) \hat{y} + z_A \hat{z}$.
Since you have written things in usual rectangular coordinates, you can take the dot product in the usual way:
$$ \begin {align*} A \cdot B &= \rho_A \rho_B \left( \cos(\phi_A) \cos(\phi_B) + \sin(\phi_A)\sin(\phi_B) \right) + z_A z_B \\ &= \rho_A \rho_B \cos(\phi_A-\phi_B) + z_A z_B \end {align*} $$
You see that $A$ and $B$ are perpendicular ($A \cdot B = 0$) if $\cos(\phi_A-\phi_B) = -\frac{z_Az_B}{\rho_A\rho_B}$.
In the example you give, you have $\rho$ and $\phi$ the same for $A$ and $B$, and $z_A = -z_B$. In this special case, the above equation for being perpendicular becomes $\rho^2 = z^2$ (since $\cos(0) = 1$).