Answer
$ y=\frac{C^{1/3}(2x+1)^{1/6}}{(x+4)^{1/9}}$.
Work Step by Step
Given $3ln(y)=\frac{1}{2}ln(2x+1)-\frac{1}{3}ln(x+4)+ln(C) $, we have $ln(y)=\frac{1}{6}ln(2x+1)-\frac{1}{9}ln(x+4)+\frac{1}{3}ln(C) \Longrightarrow ln(y)=ln(2x+1)^{1/6}-ln(x+4)^{1/9}+ln(C)^{1/3} \Longrightarrow ln(y)=ln\frac{C^{1/3}(2x+1)^{1/6}}{(x+4)^{1/9}} \Longrightarrow y=\frac{C^{1/3}(2x+1)^{1/6}}{(x+4)^{1/9}}$.
