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