<s>[INST] maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$ 
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$

Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$
How do i solve it Help me plesse
 [/INST] $K=\frac{1+t^2}{1+t+4t^2}$  To find min and max, set $K'(t)=0$ or $t^2-6t-1=0$ and solve for $t$. Answer $t=3\pm \sqrt{10}$ leading to max and min $K=\frac{20\pm 6\sqrt{10}}{80\pm 25\sqrt{10}}$. or $1.088303688022443$ and $0.245029645310883$
Corrected (stupid error in solving quadratic).
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