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

From the restriction $x^2+y^2>0$, we get that $x,y$ are not both zero.

If $x=0$, then $K={\large{\frac{1}{4}}}$.

Suppose $x\ne 0$.

Letting $t={\large{\frac{y}{x}}}$, and following your approach, we get
$$(4K-1)t^2+Kt+(K-1)=0$$
which has a real solution for $t$ if and only $K={\large{\frac{1}{4}}}$ or the discriminant
$$K^2-4(4K-1)(K-1)$$
is nonnegative.

Equivalently, either $K={\large{\frac{1}{4}}}$ or
$$-15K^2+20K-4\ge 0$$
With the restriction $K\ne{\large{\frac{1}{4}}}$, the quadratic inequality solves as
$$\frac{10-2\sqrt{10}}{15}\le K\le \frac{10+2\sqrt{10}}{15},\;\;K\ne{\large{\frac{1}{4}}}$$
Noting that 
$$\frac{10-2\sqrt{10}}{15}<\frac{1}{4}<\frac{10+2\sqrt{10}}{15}$$
it follows that 


*

*The minimum value of $K$ is ${\large{\frac{10-2\sqrt{10}}{15}}}\approx .2450296455$.$\\[8pt]$

*The maximum value of $K$ is ${\large{\frac{10+2\sqrt{10}}{15}}}\approx 1.088303688$.

