$$
\begin{split}
y' &= \frac{x^2+3y^2}{2xy} \\
&= \frac{x}{2y} + \frac{3y}{2x} \\
&= \frac{1}{2} (\frac{y}{x})^{-1} + \frac{3}{2}(\frac{y}{x})
\end{split}
$$
$$v = \frac{y}{x} \implies y = vx$$
$$
\begin{split}
y' &= v'x+ v \\
&= \frac{1}{2} v^{-1} + \frac{3}{2} v \\
\end{split}
$$
$$
\begin{split}
v'x &= y' - v \\
&= \frac{1}{2} v^{-1} + \frac{1}{2} v \\
&= \frac{1+v^2}{2v}
\end{split}
$$
$$
\begin{split}
\frac{2v}{1+v^2} dv &= \frac{1}{x} dx \\
\ln |1+v^2| &= \ln |x| + c \\
1+v^2 &= Cx, C = e^c \\
\end{split}
$$
$$
\begin{split}
1 + \frac{y^2}{x^2} - Cx &= 0\\
y^2 + x^2 - Cx^3 &= 0
\end{split}
$$
