The problem is:
Find the maximum value of $$y=\frac{x^2+x-1}{x^2-x+1}$$ assuming $x\geq 0$. I am only interested in pre-calculus solutions.
Using calculus it is easy to see that the maximum is $y=\frac{5}{3}$ occuring at $x=2$.
Other forms are
$$y=\frac{(x+\frac{1}{2})^2-\frac{3}{4}}{(x-\frac{1}{2})^2+\frac{5}{4}}$$ $$y=\frac{(\sqrt{x}-\frac{1}{\sqrt{x}})(\sqrt{x}+\frac{1}{\sqrt{x}})+1}{(\sqrt{x}-\frac{1}{\sqrt{x}})^2+1}$$
A simple approach is to rewrite it as a quadratic for $x$
$$ (y-1)x^2 - (y+1)x + (y+1)=0$$
And require that the discriminant be not negative
$$(y+1)^2 - 4(y+1)(y-1) \ge 0$$ $$(y - \frac{5}{3})(y+1) \le 0$$
Which gives us the range for $y$. To be thorough, it's also good to check that the maximum value for $y$ does not correspond to a zero in the denominator of the original fraction. In this case, it doesn't.