Answer
$ \displaystyle \frac{6x^{2}-4x -7}{(3x-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}}$
$f(x)=2x^{2}+4x+1,\quad f^{\prime}(x)=4x+4$
$g(x)=3x-1,\qquad g^{\prime}(x)=3$
$\displaystyle \frac{dy}{dx}=\frac{(4x+4)(3x-1)-(2x^{2}+4x+1)(3)}{(3x-1)^{2}}$
$=\displaystyle \frac{(12x^{2}-4x+12x-4)-(6x^{2}+12x+3)}{(3x-1)^{2}}$
$=\displaystyle \frac{6x^{2}-4x -7}{(3x-1)^{2}}$