Is there an economic way to check if an eliptic curve point in jacobian coordiate is the same as a compressed point
Computing the $z^{-1}$ is very cheap. We can do this by Euclidean algorithm, and in logarithmic time. But you can set $z=1$ in projective coordination. This make computations easy($O(1)$ since $z^{-1}=1$), and with this representation you can see the equality of points evidently.


  Note that, instead of computing $y=b*z^{-3}$ you can check  $y*z^3=b$. In this state, you don't need to finding multiplicative inverse.


Also in powerful programs such as MAGMA, $z$ is equal to $1$, by default.