Answer
$$ y'
=\frac{x(x+1)^3}{(3x-1)^2}\left( \frac{1}{ x}- \frac{3}{ x+1}-\frac{6}{3 x-1}\right).$$
Work Step by Step
Taking the $\ln $ of both sides of the equation, we get
$$\ln y= \ln \frac{x(x+1)^3}{(3x-1)^2}$$
Then using the properties of $\ln $, we have
$$\ln y= \ln x+3\ln (x+1)-2\ln(3x-1).$$
Now taking the derivative for the above equation, we have
$$\frac{y'}{y}= \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1},$$
Hence $ y'$ is given by
$$ y'=y\left( \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1}\right)\\
=\frac{x(x+1)^3}{(3x-1)^2}\left( \frac{1}{ x}+ \frac{3}{ x+1}-\frac{6}{3 x-1}\right).$$
