#### Answer

$$y=18x+1$$

#### Work Step by Step

First we will find the derivative:
$$y'=((3x-1)^{-6})'=-6(3x-1)^{-7}(3x-1)'=-6(3x-1)^{-7}\cdot3=-18(3x-1)^{-7}$$
The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula:
$$y-y_{0}=y'(x_{0})(x-x_{0})$$
For $(x_{0},y_{0})=(0,1)$ we get:
$$y-1=\left.-18(3x_{0}-1)^{-7}\right|_{x_{0}=0}(x-0)\Rightarrow y=-18(3\cdot0-1)^{-7}\cdot x+1\Rightarrow y=-18\cdot(-1)x+1\Rightarrow y=18x+1$$