Finding inverse of birational transformation.

Question

I have two algebraic curves over $\Bbb Q$:

$$C:y^2=x^4+4x^3-4$$ and the Elliptic curve

$$E:y^2=x^3 + 16x-64$$

which I computed using Sage, so that I can apply the group law on $E$ to perform arithmetic on rational points. Sage also provides a birational map $f:C\to E$ so that one can transform points on $C$ to points on $E$:

$$f:\binom xy\mapsto \binom{(4x^4 + 16x^2 + 32x)/y^2}{(8x^6 - 16x^5 - 80x^4 - 160x^3 - 64x - 64)/y^3}$$

for example, $(1,1)\in C$ maps to $(52,-376)\in E$. What I am seeking is a way to get back to $C$, i.e. the inverse of $f$. How can it be computed? The only related post I cound find was this MO post which is way above my head, and as far as I understand it's just about the existence of a rational inverse, not about its computation.

see comments by Tomita at https://math.stackexchange.com/questions/4502662/when-is-c44bc3-4b4-an-integer-square — Will Jagy, Jul 30 '22 at 17:38
There is a different elliptic curve but also no transformation, neither forward no backwards. So how han that help? — emacs drives me nuts, Jul 30 '22 at 17:42
Using that other curve $E:y^2=x^3+x-1$ , the coordinates will be smaller (divides $x$ by 4 and $y$ by 8), but the task of inverting $f$ doesn't get more obvious, IMO. @Will Jagy — emacs drives me nuts, Jul 30 '22 at 18:13
No. It's a multivariate function. Moreover, I doubt it is globally invertible so that the inverse only exists when restricted to $C$ and $E$. @Viktor Vaughn — emacs drives me nuts, Jul 31 '22 at 08:56
See MacLeod's answer [https://math.stackexchange.com/questions/1591990/birational-equivalence-of-diophantine-equations-and-elliptic-curves/1592599#1592599] — Tomita, Aug 05 '22 at 03:28

Finding inverse of birational transformation.

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