jmc

We consider both inequalities separately.

The left inequality is equivalent to $$\frac{x^2-14x+11}{x^2-2x+3}+1>0,$$or $$\frac{2x^2-16x+14}{x^2-2x+3}>0.$$Then $$\frac{x^2-8x+7}{x^2-2x+3}>0.$$The numerator factors as $$\frac{(x-1)(x-7)}{x^2-2x+3}>0.$$The denominator $x^2-2x+3=(x-1{)}^2+2$ is always positive.

The quadratic $(x-1)(x-7)$ is positive precisely when $x<1$ or $x>7.$

The right inequality is equivalent to $$1-\frac{x^2-14x+11}{x^2-2x+3}>0,$$or $$\frac{12x-8}{x^2-2x+3}>0.$$Then $$\frac{3x-2}{x^2-2x+3}>0.$$Since the denominator is always positive, this inequality holds if and only if $x>\frac{2}{3}.$

The solution is then $$x\in \boxed{\left(\frac{2}{3},1\right)\cup (7,\infty )}.$$