Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?

Without AM-GM inequality:
Let, $x=a+b$ and $y=a-b$, then you get
$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=9$$
$$14a^2+10b^2=9$$
$$x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2$$
$$x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}$$
$$x^2+y^2=\frac{9-4a^2}{5}≤\frac95$$
$$\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}}$$

If you want to find the minimum value of the expression, you can write in the same way:
$$x^2+y^2=2a^2+2b^2$$
$$x^2+y^2=2b^2+2 \times \frac{9-10b^2}{14}$$
$$x^2+y^2=\frac{4b^2+9}{7}≥\frac 97$$
Then, you get
$$\color {gold}{\boxed {\color{black} {\text{min}[x^2+y^2]=\frac 97.}}}$$
