Lagrange Multipliers with $f\left(x,y\right)=x^2-y^2$ and “constraint” $g\left(x,y\right):=2y-x^2=0$

I am working on a problem from my textbook on Lagrange Multipliers. I feel I have these down now, but I am curious about this specific problem. Let \begin{align} f\left(x,y\right)=x^2-y^2\tag{1},\\ \text{and}\;g\left(x,y\right):=2y-x^2=0.\tag{2} \end{align} The first part is easy. Since $g$ is the "constraint" as they call it, we have \begin{align} \vec{\nabla}f\left(x,y\right)&=\begin{bmatrix}2x\\-2y\end{bmatrix},\tag{3}\\\lambda\vec{\nabla}g\left(x,y\right)&=\begin{bmatrix}-2x\lambda\\2\lambda\end{bmatrix}.\tag{4} \end{align} This gives us the set of equations \begin{align} 2y-x^2&=0,\tag{5}\\ 2x&=-2x\lambda,\tag{6}\\ -2y&=2\lambda\implies \lambda=-y,\tag{7} \end{align} and substituting $\left(7\right)$ into $\left(6\right)$ gives us $y=1$, meaning $x=\pm\sqrt{2}$. Then putting these points into the original equation $\left(1\right)$ gives me $\left\{\left(\sqrt{2},1,1\right),\left(-\sqrt{2},1,1\right)\right\}$, and I have graphed to confirm.

But now what about $\left(0,0,0\right)$? It appears to be a relative minimum when I graph these two surfaces, but how do I get it to show up in my Lagrange Multipliers (unless it is not done in that way). I do notice initially that $\left(0,0,0\right)$ does satisfy both equations $\left(1\right)$ and $\left(2\right)$, however. But it does not seem this is impetus for being min alone.

Thank you for your time,

You have to make sure when solving the equations that you are not dividing by something that is zero. Having done your substitution for $\lambda$, you are left with $$y-x^2=0 \\ 2x=2xy,$$ and if $x=0$, the second equation is satisfied, and then the first equation gives $y=0$.