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 $z^2/(y-\omega^2)$.

I'd like to change your numbers slightly. (EDIT: Slight adjustment to make the formula nicer and address the correction in the comments, nothing to see here)

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 $z^7 = (1-y)(1 - \omega y)^2(1-\omega^2 y)^4$, which only ramifies over the three preimages. To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega^2 y$ lifts to an automorphism of the whole field. This automorphism maps $(1-y)(1-\omega y)^2(1-\omega^2 y)^4$ to $(1-\omega^2 y)(1- y)^2(1- \omega y)^4$, which has seventh root $z^2/(1-\omega^2 y)$.