Answer
The solution of the differential equation is given by
$$
| x| ^{3} | x^{2}-5y^{2}| = c
$$
where c is an arbitrary constant.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{x^{2}-3y^{2}}{2xy} \quad (*)
$$
Divide each term in the numerator and denominator by the highest power of $x$ , namely $x^{2}$. We obtain
$$
y^{'}=\frac{dy}{dx}=\frac{1-3(\frac{y}{x})^{2}}{2(\frac{y}{x})} =f(x,y)
$$
where $f(x,y)$ can be expressed as a function of the ratio $\frac{y}{x}$ only .
Thus eq. (*) is homogeneous.
Now we assume a new dependent variable $ v = \frac{y}{x }$
and we can express
$$
y^{'}=\frac{dy}{dx}= x \frac{dv}{dx}+v =\frac{ 1-3v^{2}}{2v}
$$
$$
x \frac{dv}{dx} = \frac{ 1-3v^{2}}{2v} -v = \frac{ 1-5v^{2}}{2v} \quad (**)
$$
The differential equation (**) is separable and can be written as
$$
\frac{2vdv}{{1-5v^{2}}} = \frac{dx}{x}
$$
integrating the left side with respect to $v$ and the right side with respect to $x$ gives
$$
\int{ \frac{2vdv}{{1-5v^{2}}} }= \int{ \frac{dx}{x}}
$$
we can obtain $v$ implicitly in terms of $ x$ as the follows
$$
\frac{-1}{5}ln| 1-5v^{2} |= ln | c x |
$$
where c is an arbitrary constant.
$$
| 1-5v^{2}| = c | x| ^{-5}
$$
we find the solution of Eq. (*) by replacing $v$ by $ y/x $ in the previous expression,
$$
| x| ^{5} | 1-5(\frac{y}{x})^{2}| = c
$$
i.e.
$$
| x| ^{3} | x^{2}-5y^{2}| = c
$$
where c is an arbitrary constant.
