subject to $x^3+3x^2y^2+3y^4=7$.
I write the Lagrangian as $L(x,y,\lambda)=x+y^2-\lambda(x^3+3x^2y^2+3y^4-7)$ and partially differentiate each variable:
$L'_x(x,y,\lambda)=1-3\lambda x^2-6\lambda x y^2$
$L'_y(x,y,\lambda)=2y-6\lambda x^2y-12\lambda y^3$
which gives me the system:
1) $1-3\lambda x^2-6\lambda x y^2=0$
2) $2y-6\lambda x^2y-12\lambda y^3=0 \Rightarrow 1-3\lambda x^2-6\lambda y^2=0$
Then I equate equation 1) and 2) which solves for $x=1$. Plugging $x=1$ back into equation 3) gives me $y=1$ or $y=-1$.
The question is, how can the Lagrangian have 3 stationary points in this case? it should be evident that without solving for $\lambda$ that I only get two stationary points:
$(1,1,\lambda)$ and $(1,-1,\lambda)$
Yet, there should be three stationary points, what do I miss?