Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$. 
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$

My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that 
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$ 
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so that $y^2\le \frac{4}{7}$ and $x^2\le \frac{8}{7}$.
Thus $x^2+2y^2\le \frac{8}{7}+\frac{8}{7}=\frac{16}{7}$.
But clearly $x^2+2y^2\neq \frac{16}{7}$, since if $\begin{cases}x^2=\frac{8}{7}\\y^2=\frac{4}{7}\end{cases}$, then $x^2-xy+2y^2\neq 1$  
so that equality can't be achieved and we only have that $x^2+2y^2<\frac{16}{7}$.

You already have a great answer from Kim.  Another way would be to use Cauchy-Schwarz inequality to get the same bounds on $xy$:
$$(1+xy)^2 = (x^2+2y^2)(2y^2+x^2) \ge 8x^2y^2$$
$$\implies -1-xy \le 2\sqrt2 xy \le 1+xy$$
$$\implies -\frac1{2\sqrt2+1} \le xy \le \frac1{2\sqrt2-1}$$
$$\implies 1-\frac1{2\sqrt2+1}\le 1+xy=x^2+2y^2 \le 1+\frac1{2\sqrt2-1}$$
