Rational map of a curve to an elliptic curve

I think that taking the quotient amounts to the following. I think about this at the level of function fields. So let $k$ be the field of constants. Then the function field of your curve is $F=k(x,y)$, which is a quadratic extension of the field of rational functions $K=k(x)$, where $y^2=(x^3-1)(x^3-a)$.

The symmetry that you talk about can be viewed as an automorphism $\phi$ of the field $F$. To avoid fractional exponents I denote $a=A^6$. Here I need to assume that $a$ has a sixth root in $k$. Hopefully this will not pose a problem. Now we can define the $k$-automorphism of $F$ by the rules $$ \phi:x\mapsto\frac{A^2}x,\qquad y\mapsto \frac{yA^3}{x^3}. $$ We need to verify that this is an automorphism of $F$. This is equivalent to checking that $\phi(y)^2=(\phi(x)^3-1)(\phi(x)^3-A^6)$, and verifying that is straightforward.

You hinted at the possibility that we are moding out an involution. This is the case, because we easily see that $$ \phi(\phi(x))=x,\qquad\text{and}\qquad\phi(\phi(y))=y. $$ Thus $\phi$ generates a cyclic group of order two. Basic Galois theory then tells us that $L=\operatorname{Inv}(\phi)$ is a subfield of $F$ such that $[F:L]=2$. Our next task is to identify the fixed field $L$. The first and most obvious observation is that $$ u=x+\phi(x)=x+\frac{A^2}x\in L. $$ This is immediate from the involutory property of $\phi$. At this point we observe that $x$ is algebraic of degree two over the field $M=k(u)$. Thus we conclude that $M=L\cap K$.

Finding another "useful" element in $L$ was a bit tricky. It is easy enough to find elements from $L\setminus M$, but they were kludgy to work with. A choice that worked for me is $$ v=\frac yx+\phi(\frac yx)=y\left(\frac1x+\frac A{x^2}\right). $$ Clearly $v^2$ will be an element of $M$ - we just need to calculate it $$ \begin{aligned} v^2&=\frac{y^2}{x^2}\left(1+\frac Ax\right)^2=\frac{(x^3-1)(x^3-A^6)}{x^2}\left(1+\frac Ax\right)^2\\ &=\frac{A^8 + 2 A^7 x + A^6 x^2 - A^2 x^3 - A^8 x^3 - 2 A x^4 - 2 A^7 x^4 - x^5 - A^6 x^5 + A^2 x^6 + 2 A x^7 + x^8}{x^4}\\ &=u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A).\\ \end{aligned} $$

It is easy to see that $L=M(v)=k(u,v)$. Evidently $k(u,v)\subseteq L$. Also clearly $y\in k(u,v,x)$, so $F=k(u,v,x)$ and we saw that $x$ is quadratic over $k(u,v)$ so $[F:k(u,v)]=2$. As $[F:L]=2$ the claim follows.

From the general theory of hyperelliptic curves we infer that the equation $$ v^2=u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A)\qquad(*) $$ defines a curve of genus $g=1$ whenever the roots of that quartic are distinct. Therefore $L$ is the function field of an elliptic curve whenever that holds.

The process of transforming the curve $(*)$ to a cubic needs to move one of the zeros of the r.h.s. to the infinity. Here we observe that $u=-2A$ is a zero of that quartic, so let's work with that. We can factor $$ u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A)=(u+2A)(u^3-3A^2u-A^6-1). $$ Here by Taylor expansion around $u=-2A$ $$ u^3-3A^2u-A^6-1=(u+2A)^3-6A(u+2A)^2+9A^2(u+2A)-(A^3+1)^2. $$ This implies that dividing $(*)$ by $(u+2A)^4$ gives us, in the variables $$ w=\frac{v}{(u+2A)^2},\qquad z=\frac1{u+2A}, $$ the equation $$ w^2=1-6Az+9A^2z^2-(A^3+1)^2z^3.\qquad(**) $$ Getting warmer! To get to the Weierstrass form we multiply $(**)$ by $(A^3+1)^4$ and introduce, finally, the variables $$ s=(A^3+1)^2w,\qquad t=-(A^3+1)^2z, $$ whence the equation takes the form $$ s^2=t^3+9A^2t^2+6A(A^3+1)^2z+(A^3+1)^4. $$