Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$

At the extreme points we must have
\[\nabla f= \lambda \nabla g,\]
where $g=x^2+2y^2.$ Hence
\[(1,0)=(2\lambda x,4 \lambda y).\]
So $4 \lambda y=0$ and  $2 \lambda x=1$.  So $\lambda \neq 0$, and from the fist equation we get
\[y=0.\]
Substituting $y=0$ in the constraint, we get
\[x^2=2 \Rightarrow x= \pm\sqrt{2}.\]
We have 
$$f(\sqrt{2},0)=\sqrt{2},    f(-\sqrt{2},0)=-\sqrt{2}.$$
So $(\sqrt{2},0)$ is the maximum and $(-\sqrt{2},0)$ is the minimum point.