Question: Given below

$2y^2 + x^2$ 
Such that $x + y =1$
 How do I show that it is a minimum or maximum point. 

What I have tried:
I differentiated wrt x

Answer: 
Since $x+y=1$, $2y^2+x^2=2y^2+(1-y)^2=3y^2-2y+1$. If $f(y)=3y^2-2y+1$, then $f'(y)=6y-2$. So, the minimum is attained when $y=\frac13$, and when that happens $x=\frac23$.
I suppose $x,y \in \mathbf{R}$. There is no maximum since if $n$ is a sufficiently large integer then $n+(1-n)=1$ and $2(1-n)^2+n^2 \to \infty$. About the minimum, if $x=1-y$ then
$$
2y^2+x^2=3y^2-2y+1=3\left(y-\frac{1}{3}\right)^2+\frac{8}{9} \ge \frac{8}{9}
$$
with equality iff $y=\frac{1}{3}$ and $x=\frac{2}{3}$.