Answer
$\displaystyle \frac{(4x-4)(x^{2}+x+1)-(x^{2}-4x+1)(2x+1)}{(x^{2}+x+1)^{2}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=x^{2}-4x+1,$
$f^{\prime}(x)=2(2x)-4=4x-4$
$g(x)=x^{2}+x+1,$
$g^{\prime}(x)=2x+1$
$\displaystyle \frac{dy}{dx}=\frac{(4x-4)(x^{2}+x+1)-(x^{2}-4x+1)(2x+1)}{(x^{2}+x+1)^{2}}$
... we do not need to expand the answer..