For all real x, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

$\frac{1}{3}$

$y=\frac{1-x+x^2}{1+x+x^2}=1-\frac{2 x}{1+x+x^2}=1-\frac{2}{\frac{1}{x}+1+x}=1-\frac{2}{t}$

where $t=\frac{1}{x}+1+x$

Now, y is min : when $\frac{2}{t}$ is max. ⇒ t is min.

∴ $\frac{d t}{d x}=-\frac{1}{x^2}+1=0 \Rightarrow x= \pm 1$

$\frac{d^2 t}{d x^2}=\frac{2}{x^3}>0$  for x = 1

∴  min. value of y is $1-\frac{2}{1+1+1}=1-\frac{2}{3}=\frac{1}{3}$.