If $x+y+z=0, \quad \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=?$

x + y + z = 0 $\Rightarrow(x+y+z)^{2}=0$ $\Rightarrow x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=0$ $\Rightarrow x^{2}+y^{2}+z^{2}=-2(x y+y z+z x)$ On squaring both sides, (x² + y² + z²)² = 4 (xy + yz + zx)² $\left(x^{2}+y^{2}+z^{2}\right)^{2}=4(x y+y z+z x)^{2}$ $=4\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+2 x y^{2} z+2 x y z^{2}+2 x^{2} y z\right)$ $\begin{aligned} \Rightarrow & x^{4}+y^{4}+z^{4}=2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+2 x y z(x+y\\ &+z) \end{aligned}$ $\begin{aligned} \Rightarrow & x^{4}+y^{4}+z^{4}=2\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right) \\ &[\because x+y+z=0] \end{aligned}$ $\Rightarrow \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=\frac{1}{2}$

Alternative:

Given : x + y + z = 0

Let x = 0

y = 1

z = –1

According to the question

$\Rightarrow \frac{x^{y} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{4}+y^{4}+z^{4}}=\frac{0+1+0}{0+1+1}=\frac{1}{2}$