z-transform causality properties: negative coefficents are zero ($x[-1]z^1=0$, $x[-2]z^2=0$, ...) Let's suppose I have a system: 

$$Y(z)=X(z)H(z)$$

If the system is causal, does that mean that all the negative coefficients (example: x[-1]) of the transform for $Y(z)$, $X(z)$, and $H(z)$ are zero?  or does that rule apply only to $H(z)$?  what about $Y(z)$ and $X(z)$?
No. The system response is usually $H(z)$. If $H$ is causal, then it has no non-causal coefficients.  $X$ may very well be a signal that exists for all time (and so existed before time 0) and, therefore, have non-causal coefficients.  That means $Y$ will also have non-causal coefficients.

If $Y$ is taken as the system and $Y$ is causal, then your statement is true. But usually $Y$ is the output, $X$ is the input, and $H$ is the system response.