Riemann surfaces exercise I am trying to do an exercise from Rick Miranda's Book 

Let $U$ be the affine plane curve defined by $x^2=3+10t^4+3t^8$ and $V$ defined by $w^2=z^6-1$, show that the function $F :U \rightarrow V$, $$F(x,t)=\left(\frac{1+t^2}{1-t^2},\frac{2tx}{(1-t^2)^3}\right)$$ is holomorphic and nowhere ramified when $t \neq  \pm 1$. 

Well to prove that this is holomorphic i simply separated in $4$ cases for the coordinate charts that we could have and checked that the local representation of these functions were holomorphic. To prove that part about the unramified points and i was trying to check the derivative of these representations and see where they would be zero, hopefully nowhere in the surface would be the result we wanted, however using this and the fact that we know the derivative of the implicit function i was not able to prove this, so my question is if this is the right way to think abou this problem, or should i be taking another approach, should i try to explicit right with are the implicit functis envolved in this problem? Thanks in advance.
New edit:
I was able to prove this was true for two of  the representations , now im trying to prove it for the other two, where in one we suppose that $t$ is given by $h(x)$ and $z$ is given by $k(w)$, and the other one where $x$ is given by $j(t)$ and $z$ is given by $k(w)$. 
Calculating the maps and their derivatives i cant seem to get a contradiction for why the derivative cannot be zero.
At this point what i cant seem to do is this

So suppose now the charts is $t=h(x)$, so $\frac{df}{dt} \neq 0$ and $z$ depends on $w$ by some holomorphic function k,so $\frac{dk}{dw}\neq 0$, we have  in local coordinates $\psi_2 \circ f \circ \phi_1(x) = \frac{2h(x)x}{(1-h(x)^2)^3}$ with derivative $\frac{2(5xh(x)^2h'(x)+xh'(x)-h(x)^3+h(x))}{(1-h(x)^2)^4}$ and i cant seem to find a contradition for why this derivative cannot be zero using the fact that those derivatives cannot be zero, so any tips on something i may be forgetting would be helpfull.

 A: The map $F\colon U \rightarrow V$ is holomorphic if so as a map $F\colon U \rightarrow \mathbb{C}^2$. Then we may only check on a local parameterization for $U$. It follows that $F$ is holomorphic everywhere except for the four points $(\pm 4, \pm 1)\in U$ where it is not defined.
Now to compute the ramification locus we may use the equations. Indeed we will use the pullback of differential forms. 
Let $C,D$ be two affine plane curves is given respectively by the equations $f(x,y)=0$ and $g(x,y)=0$. Let $F\colon D \rightarrow C$ be a holomorphic map. For a point $p\in D$, a vector $v\in T_pD$ and for a generator of $\omega_{F(p)}$ of the cotangent space $T_{F(p)}^\ast C$ we have that
$$
\omega_{F(p)}(DF_pv) = 0 \Longleftrightarrow DF_pv = 0.
$$
Then the ramification locus is given by the zeros of $F^\ast \omega$ for a local one-form without zeros. The advantage of this approach is that for plane curves we may use that the cotangent bundle of $C$ is generated by the classes of $dx,dy$ modulo $df$. Then
we may compute the common zeros of 
$$
((F^\ast dx)\wedge dg)|_D, \quad ((F^\ast dy)\wedge dg)|_D
$$
as from the implicit function theorem, $x$ or $y$ will be a local parameter for $C$ so $dx$ or $dy$ will restrict to a form without zeros on $C$.
Now we will apply this to our example. Let $f = w^2-z^6+1$ define $V$, let $g= x^2-3-10t^4-3t^8$ define $U$ and let $$(z,w) = F(x,t) = \left(\frac{1+t^2}{1-t^2}, \frac{2tx}{(1-t^2)^3}\right)$$
We have that 
$$
(F^\ast dz)\wedge dg = d\left(\frac{1+t^2}{1-t^2}\right) \wedge dg = \frac{8xt}{(1-t^2)^2}dt\wedge dx, \\
(F^\ast dw)\wedge dg = d\left(\frac{2tx}{(1-t^2)^3}\right) \wedge dg =\frac{4(12t^{10}-12t^8+20t^6-20t^4-5t^2x^2-x^2)}{(t^2-1)^4}dx\wedge dt
$$
Restricting to $U$ gives 
$$
(F^\ast dz)\wedge dg = d\left(\frac{1+t^2}{1-t^2}\right) \wedge dg = \frac{8xt}{(1-t^2)^2}dt\wedge dx, \\
(F^\ast dw)\wedge dg = d\left(\frac{2tx}{(1-t^2)^3}\right) \wedge dg =\frac{-12(t^2+1)^5}{(t^2-1)^4}dx\wedge dt
$$
and it is clear that they do not share a common zero. Indeed for $t = \pm i$ in $U$ we must have $x= \pm 4$.
ADDED: This was a more direct approach using the extrinsic geometry as they are plane curve we don't need to use charts of parameterizations. 
From just the definition we may use that the projection $(z,w) \mapsto z$, restricted to $V$,  define a coordinate chart on the open set given by $\dfrac{\partial f}{\partial w} \neq 0$ and the same holds for $(z,w) \mapsto w$ and $\dfrac{\partial f}{\partial z} \neq 0$. Then a local expression for $F$ is
$$
z \circ F \circ \gamma \quad \text{or} \quad w \circ F \circ \gamma
$$
where $\gamma$ is a local parameterization for $U$. Therefore we may check the orders of the functions
$$
\frac{1+t^2}{1-t^2} \quad \text{and} \quad \frac{2tx}{(1-t^2)^3}
$$
where $t= t(x)$ or $x=x(t)$. For $t=t(x)$ and $\phi = w\circ F = \frac{2tx}{(1-t^2)^3}$ we have from $x^2 =3+10t^4+3t^8$ that
$$
t'= \frac{x}{20t^3+12t^7}
$$
Then 
$$
\phi'(x) = 2\frac{5t^2t'x+xt'-t^3+t}{(1-t^2)^4} = \\ = \frac{5t^2\left(\frac{x}{20t^3+12t^7}\right)x+x\left(\frac{x}{20t^3+12t^7}\right)-t^3+t}{(1-t^2)^4} = \frac{5t^2x^2 + x^2 + (-2t^3+2t)(10t^3+6t^7)}{(10t^3+6t^7)(1-t^2)^4} = \\ = \frac{3(t^2+1)^5}{(10t^3+6t^7)(1-t^2)^4}
$$
where we have used again $x^2 =3+10t^4+3t^8$. Then the only possible ramification points are $t= \pm i$ i.e. $(x,t) = (\pm 4, \pm i)$. However $\phi$ does not give a local representative for $F$ around these points. Indeed 
$$
F(\pm 4 , \pm i) = \left(0, \pm 1\right) 
$$
and around these points $w$ does not give a chart since $\dfrac{\partial f}{\partial z} (0,\pm 1)= 0$.
It remains to make this calculation for the other three representatives.
