<s>[INST] Solving system of equation using Macaulay2 As an algebraic curve, the Klein quartic can be viewed as a projective algebraic curve over the complex numbers $\mathbb{C}$, defined by the following quartic equation in homogeneous coordinates $[x:y:z]$ on $\mathbb{P}^2_{\mathbb{C}}$:
$$x^3 y + y^3 z + z^3 x = 0.$$
Now we want to find the eigenvectors of this curve, therefore consider the following matrix:
$$
 \begin{bmatrix}
 3x^2y+z^3 & x^3 + 3y^2z & y^3+3z^2  \\
 x & y & z  \\
 \end{bmatrix},
 $$
after computing the $2$-minors we got the following system:
\begin{equation*}
\left\{
\begin{alignedat}{3}
% R & L   &  R & L   &  R & L 
3x^2y^2+z^3y & -{} &  (3y^2zx + x^4) &  = 0 \\
3z^2x^2+y^3x & -{} &  (3x^2yz + z^4)& = 0 \\
3y^2z^2+x^3z & -{} &  (3z^2xy + y^4)& = 0 
\end{alignedat} \ .
\right.
\end{equation*}
I am new to Macaualy 2 and I am wondering if Macaulay2 can help me to solve the above system numerically with loading NumericalAlgebraicGeomety package, (I know that because of infinity many solution we need normalize the system on unit sphere).
 [/INST] I divided by $z^4$ all equations and set $$\frac{x}{z}=X;\;\frac{y}{z}=Y$$
so I got
$$
\begin{cases}
-X^4+3 X^2 Y^2-3 X Y^2+Y=0\\
-3 X^2 Y+3 X^2+X Y^3-1=0\\
X^3-3 X Y-Y^4+3 Y^2=0\\
\end{cases}
$$
solved by Mathematica I got these solutions
$$
\begin{array}{rr}
 X & Y\\
\hline
 1 & 1 \\
 0.307979 & 1.55496 \\
 0.643104 & 0.198062 \\
 5.04892 & 3.24698 \\
 -1.80194 & 0.801938 \\
 -0.445042 & -0.554958 \\
 1.24698 & -2.24698 \\
 -0.610856 & 0.106042 \\
 -0.173596 & -1.63705 \\
 0.667631 & 1.26654 \\
 0.789553 & 0.52713 \\
 1.89706 & 1.49783 \\
 9.4302 & -5.76459 \\
\end{array}
$$
</s>