Let \begin{align*} &\boldsymbol{e}_0=-x_{{0}}{t}^{3}-x_{{1}}t+{\frac {x_{{2}}}{t}}+{\frac {x_{{3}}}{{t}^{3} }},\\ &\boldsymbol{e}_1=x_{{0}}{t}^{3}-x_{{1}}t-{\frac {x_{{2}}}{t}}+{\frac {x_{{3}}}{{t}^{3}} } ,\\ &\boldsymbol{e}_2=x_{{0}}{t}^{3}+3\,x_{{1}}t+3\,{\frac {x_{{2}}}{t}}+{\frac {x_{{3}}}{{t }^{3}}} ,\\ &\boldsymbol{e}_3-x_{{0}}{t}^{3}+3\,x_{{1}}t-3\,{\frac {x_{{2}}}{t}}+{\frac {x_{{3}}}{{ t}^{3}}} \end{align*}
I need to eliminate the variable $t$, that is, I am looking for a homogeneous polynomial $f(y_0,y_1,y_2,y_3)$ such that $$ f(\boldsymbol{e}_0, \boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3)= g(x_0,x_1,x_2,x_3), (*) $$ and the polynomial $g$ does not depend on $t$.
Furthermore, for other considerations, it is known that the quantity $\alpha_1+2 \alpha_2+3 \alpha_3$ is the same for all monomials $y_0^{\alpha_0} y_1^{\alpha_1} y_2^{\alpha_2} y_3^{\alpha_3}$ of the polynomial $f(y_0,y_1,y_2,y_3)$.
My attempts.
By brute force, I have found the following polynomial $$ f(y_0,y_1,y_2,y_3)={y_{{0}}}^{2}{y_{{3}}}^{2}-6\,y_{{0}}y_{{1}}y_{{2}}y_{{3}}+4\,y_{{0}}{ y_{{2}}}^{3}+4\,{y_{{1}}}^{3}y_{{3}}-3\,{y_{{1}}}^{2}{y_{{2}}}^{2}. $$
Upon substituting $y \mapsto e$ we obtain $$ f(\boldsymbol{e}_0, \boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3)=64 f(x_0,x_1,x_2,x_3). $$
I am dissatisfied with the approach because it is brute force and cannot be applied to other similar problems.
Question 1. Is there a comprehensible unified algorithm for the elimination of the variable $t$ from the equation that could be applied to similar tasks?
Question 2. Are there other polynomials that eliminate $t$, apart from the powers of $f$?
Let $\mathbf{\lambda} = (\lambda_0,\lambda_1,\lambda_2,\lambda_3)^{\intercal} = (3,1,-1,-3)^{\intercal}$, let $$ C = \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ -3 & -1 & 1 & 3 \\ 3 & -1 & -1 & 3 \\ 1 & 1 & -1 & 1 \end{array}\right), \quad \text{ and } \quad \mathbf x(t) = \left(\begin{array}{c} x_0 t^3 \\ x_1 t \\ x_2/t \\ x_3/t^3\end{array} \right). $$ Then if $\mathbf e = (e_0,e_1,e_2,e_3)^{\intercal}$ we have $\mathbf e(t) = C\mathbf x(t)$. Now $C$ is invertible with $$ C^{-1}= \frac{1}{8} \left( \begin{array}{cccc} 1 & -3 & 3 & 1 \\ 1 & -1 & -1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & 3 & 3 & 1 \end{array} \right) $$ hence $\mathbf x(t) = C^{-1} \mathbf e(t) = (y_0(\mathbf e), y_1(\mathbf e), y_2(\mathbf e), y_3(\mathbf e))^\intercal$. More explicitly, we have $$ \begin{split} y_0(t) &= \frac{1}{8}\left(e_0-3e_1+3e_2-e_3 \right) = x_0 t^3; \\ y_1(t) &= \frac{1}{8}\left(e_0-e_1-e_2+e_3 \right) = x_1 t; \\ y_2(t) &= \frac{1}{8}\left(e_0+e_1-e_2-e_3 \right) = x_2 t^{-1}; \\ y_3(t) &= \frac{1}{8}\left(e_0+3e_1+3e_2 +e_3\right) = x_3 t^{-3}. \end{split} $$
Now because $C$ is invertible, the algebra of functions $\mathbb C[e_0,e_1.e_2,e_3]$ is equal to $\mathbb C[y_0,y_1,y_2,y_3]$. But a polynomial in $\{y_0,y_1,y_2,y_3\}$ is independent of $t$ if and only if each monomial which occurs with nonzero coefficient in it is, and the monomial $\prod_{j=0}^3 y_j^{a_j}$ is independent of $t$ if and only if $3a_0 + a_1 = a_2+3a_3$. It is easy to see from this that any such monomial is a product of powers of $$ f_0 = y_0y_3, f_1= y_1y_2, f_2 = y_0y_2^3, f_3 = y_1^3y_3, $$ that is, the subring of $\mathbb C[e_0,e_1,e_2,e_3]$ consisting of the elements which are independent of $t$ is $\mathbb C[f_0,f_1,f_2,f_3]$.