I'd like to change your numbers slightly.
One solution is to set $y^3 = x$, triply ramified only over $0$ and $\infty$, and if we want the 7-fold ramification over $x=1$ (which has solutions $y=1,\omega,\omega^2$) we set $z^7 = (y-1)(y-\omega)^2(y-\omega^2)^4$, which only ramifies over the three preimages. To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega y$ lifts to an automorphism of $\mathbb{C}(x,y,z)$. This automorphism maps $(y-1)(y-\omega)^2(y-\omega^2)^4$ to $(y-\omega^2)(y-1)^2(y-\omega)^4$, which has seventh root $z^2/(y-\omega^2)$.