Updated this question, just focusing on the relevant part
$f_2(u,v) = \tfrac{27}{64} u^3 - \tfrac{81}{64} u^2 v + \tfrac{189}{20} u^2 + \tfrac{81}{64} u v^2 - \tfrac{189}{10} u v + \tfrac{1764}{25} u + \tfrac{165}{64} v^3 - \tfrac{99}{4} v^2 + \tfrac{297}{5} v$
f2(u,v) = 27/64*u^3 + (-81/64*v + 189/20)*u^2 + (81/64*v^2 - 189/10*v + 1764/25)*u + (165/64*v^3 - 99/4*v^2 + 297/5*v)
$f_3(s,t) = s^2 + \tfrac{1485}{16} t^3 + \tfrac{2673}{16} t^2 + \tfrac{8019}{80} t + \tfrac{11907}{400}$
f3(s,t) = 1485/16*t^3 + 2673/16*t^2 + 8019/80*t + (s^2 + 11907/400)
Given the substitution
$t = \frac{v}{u}, s = \tfrac{1}{160 u^2}\left[135 u^2 - 405 u^2 v + 1512 u^2 + 405 u v^2 - 3024 u v + 825 v^3 - 3960 v^2\right]$
t = v/u s = (135*u^3 + (-405*v + 1512)*u^2 + (405*v^2 - 3024*v)*u + (825*v^3 - 3960*v^2))/(160*u^2)
I have checked that $f_2(u,v) = \frac{16 u}{165 t^3 + 81 t^2 - 81 t + 27} f_3(s,t)$
also we have the inverse
u = (3960*t^2 + 3024*t + (160*s - 1512))/(825*t^3 + 405*t^2 - 405*t + 135) v = t*u
So in what way are the rational solutions of $f_2$ related to those of $f_3$? I don't think they are in bijection because of the problem of zero denominators. Although, it does seem that maybe I can get around it since $165 t^3 + 81 t^2 - 81 t + 27$ is not zero for any rational $t$.. is that the general way to deal with it? I wonder what happens if the denominator had a rational root. On the other hand, if $u=0$ then the LHS might is not zero (except for $v=0,24/5$) but the RHS is...
old version of the question, ignorable:
I have started with the elliptic curve $f$:
$f(x,y) = x^3 + 3y^3 - 11$
and using the techniques here transformed it into the Weierstrass form
$w(s,t) = s^2 + 37125 t^3 + 66825 t^2 + 40095 t + 11907$
using the change of variables
$ \begin{eqnarray} s &=& \frac{360 x^3 - 3024 x^2 + 1080 y^3 + 6156 y^2 + 1980}{16 x^2 + 24 x y - 88 x + 9 y^2 - 66 y + 121} \\ t &=& \frac{15 y + 57}{20 x + 15 y - 55}. \end{eqnarray}$
The problem, we don't have $f(x,y) = w(s,t)$, instead we have the identity $ \frac{f(x,y)}{4x+3y-11} = \frac{w(s,t)}{12375 t^3 + 6075 t^2 - 6075 t + 2025}.$
So my question is how can I be sure that I have found a birational transformation? If I want to study the rational solutions of $f$ via $w$ I need to know the exact relation between the two and how to avoid problems related to zero denominators. Thanks very much.