How can we solve the following system? \begin{align*}&y-3x^2y-y^3+2\lambda x=0 \\ & x-x^3-3xy^2+2\lambda y =0 \\ & x^2+y^2-1=0 \end{align*} I have done the following:
Multiply the first equation by $y$ and the second by $x$ : \begin{align*}&y^2-3x^2y^2-y^4+2\lambda xy=0 \\ & x^2-x^4-3x^2y^2+2\lambda xy =0 \\ & x^2+y^2-1=0 \end{align*} Second equation minus first equation : \begin{align*}&y^2-3x^2y^2-y^4+2\lambda xy=0 \\ & x^2-x^4-y^2+y^4 =0 \\ & x^2+y^2-1=0 \end{align*} From last equation we solve for $x^2$ : \begin{align*}&y^2-3x^2y^2-y^4+2\lambda xy=0 \\ & x^2-x^4-y^2+y^4 =0 \\ & x^2=1-y^2\end{align*} Substitute in the first equation the term $x^2$ : \begin{align*}&y^2-3x^2y^2-y^4+2\lambda xy=0 \\ & 1-y^2-(1-y^2)^2-y^2+y^4 =0 \\ & x^2=1-y^2\end{align*} The second equation is a trivial one, right?
What else could we do to find the solutions?