Answer
$f^{\prime}(x)=-3(3x-1)^{-2}$
Work Step by Step
Let $u(x)=\displaystyle \frac{1}{x}=x^{-1},\qquad v(x)=3x-1$
$\displaystyle \frac{du}{dx}=-x^{-2}, \quad \frac{dv}{dx}=3$
Then,$\quad y=f(x)=u(v(x)$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-v^{-2}=-(3x-1)^{-2}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=$
$-(3x-1)^{-2}\cdot 3=-3(3x-1)^{-2}$
