Minimizing $f(x,y,z)=y$ In the problem of minimizing $f(x,y,z)=y$ over the constraint set $z=y^3 - x^2$ and $z=x^2$, I have managed to solve the problem directly and obtain that the minimum occurs at $x=0, y=0, z=0$, yielding a value of $f(0,0,0)=0$, but when I write the Lagrangian and try to solve it, I realize that it has no solutions. I am trying to understand why the Lagrange multipliers didn't work here and any insights on this would be helpful.
Edit: $\nabla f+λ_1∇g_1+λ_2∇g_2=0$ yields: $x:−2xλ_1+2xλ_2=0$ $y:1+3y^2λ_1=0$ $z:−λ_1−λ_2=0$ Which gives: $λ_1=−λ_2$ and $λ_2x=0$ but since $λ_2$ cannot be $0$, $x=0$, which gives $y^3−z=0$ and $−z=0$, giving $y=0$ and $z=0$, but $1+3y^2λ_1=0$ is not satisfied.
Thank you for your help!

Let $g(x,y,z):=(z-y^3+x^2, z-x^2)$, $h(x,y,z, \lambda, \mu):=f(x,y,z)+\lambda(z-y^3+x^2)+ \mu(z-x^2)$ and $\phi:= \nabla h$.
Lagrange says: if(!) $f$ has in $(x_0,y_0,z_0) $ a local extremum under the constraints $g(x,y,z)=(0,0)$ and if(!) $rank g'(x_0,y_0,z_0)=2$, then there are $\lambda_0, \mu_0$ such that $\phi(x_0,y_0,z_0, \lambda_0, \mu_0)=0$.
In your case we have $x_0=y_0=z_0=0$, but $rank g'(0,0,0)=1 \ne 2$.
