Compare squared and reflected inputs

Subtract 1 to obtain $\dfrac{x^2-x+1}{-(x+1)(x-3)}<0$. The numerator $x^2-x+1=(x-\frac12)^2+\frac34$ is always positive. Alternatively, its discriminant is $1-4=-3<0$ and its leading coefficient is positive. Thus $\dfrac1{(x+1)(x-3)}>0$, with signs $+,-,+$ across the excluded poles $-1,3$. Hence $x<-1$ or $x>3$.

Replace the original input by $x^2$. Then $x^2<-1$ or $x^2>3$. Since $x^2\geq0$, reject the first branch. Thus $x<-\sqrt3$ or $x>\sqrt3$.

Replace the original input by $-x$. The fraction becomes $\dfrac{-x+4}{-x^2-2x+3}=\dfrac{x-4}{x^2+2x-3}$. Thus $-x<-1$ or $-x>3$, giving $x>1$ or $x<-3$.