Lagrange multiplier determinant condition in two variables ($f_x{g}_y-f_y{g}_x=0$)

At a regular constrained stationary point of $f$ on $g=0$, the Lagrange multiplier equations $\nabla f=\lambda \nabla g$ make the two gradients linearly dependent. Eliminating $\lambda$ gives the displayed zero determinant. This condition is necessary, not sufficient for an extremum. At a singular constraint point where $\nabla g=0$, the determinant is automatically zero even when no multiplier exists; for example $g=x^2+y^2$ and $f=x$ at the origin.