Find minimum of piecewise defined function

Case-$1$: $x^2+2y^2-1\leq0$. Problem simplifies to: $$\ \min_{(x, y) \in \mathbb{R}^2} x \quad s.t. \quad x^2+2y^2-1\leq0 $$ It's easy to see that the minimum value is $-1$ and occurs at $(-1, 0)$. However, you can use Lagrange Multiplier to solve this as done for the case-2 below.

Case-$2$: $x^2+2y^2-1\geq0$. Problem simplifies to: $$\ \min_{(x, y) \in \mathbb{R}^2} x+10(x^2+2y^2-1) \quad s.t. \quad x^2+2y^2-1\geq0 $$ Writing Lagrangian for the problem: $$\ L(x, y, \lambda)=x+10(x^2+2y^2-1) - \lambda(x^2+2y^2-1) \quad;\lambda\geq0 $$ Setting $\frac{\partial L}{\partial x}=0$ and $\frac{\partial L}{\partial y}=0$ we get, $$\ x=\frac{1}{2\lambda-20}\text{ and } y=0 $$ Substituting it back in the constraint, we get $\lambda=\frac{19}{2}$ or $\lambda=\frac{21}{2}$. Finally, you can plug it back and verify that $\lambda=\frac{19}{2}$ corresponds to the global minimum giving us $(-1, 0)$ as the solution.

In case if you are wondering what happens to $\lambda=\frac{21}{2}$. Its a local maximum.

Note: You can argue directly that if constraint is inactive i.e. $x^2+2y^2-1>0$. There exists no optimal for the objective and if the constraint is active, case-2 and case-1 are exactly same.