Find the extreme values of $f(x, y)=x^2+2y^2$ on the circle $x^2+y^2=1$ using Lagrange's multiplier method.
My attempt:
The constraint function $g$ is $g(x, y)=x^2+y^2-1$. So $\vec\nabla f=\lambda\vec\nabla g$ implies that $$2x\vec i+4y\vec j=\lambda(2x\vec i+2y\vec j)\Rightarrow 2x=2x\lambda, 4y=2y\lambda.$$ Now from these, either $x=0, y=0$ which is not a possible solution because it is not on $g(x, y)$. But after this, I could not go further. What should I do next?