$$
\frac{dy}{dx} = \frac {x^2-3y^2}{2xy}
$$
It’s not homogeneous, since: $$\frac{dy}{dx}*2xy - x^2 + 3y^2 = 0 $$
Using substitution to solve the equation:
Let $$u = \frac {y}{x} \\then, y = ux$$
$$\frac {dy}{dx} = \frac {du}{dx} * x + u$$
and,
$$\frac {dy}{dx} = \frac {x^2-3y^2}{2xy} \\ = \frac {1-3(y/x)^2}{2(y/x)} \\ = \frac {1-3(u^2)}{2u} $$
then,
$$\frac {1-3(u^2)}{2u} = \frac {du}{dx} * x + u \\
\frac {1-3(u^2)}{2u} - u = \frac {du}{dx} * x \\
\frac {1-5(u^2)}{2u} = \frac {du}{dx} * x \\
\frac{2u}{1-5(u^2)} du = \frac {1}{x} dx $$
integrating both sides:
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx$$
Using substitution again: $$let \ a = 1-5(u^2),$$
$$\frac{da}{du} = -10u => du = \frac{da}{-10u}$$
$$\int\frac{2u}{1-5(u^2)} du = \int\frac{1}{-5a} da \\= -\frac{1}{5}\ln{|a|}+C $$
$$ => \int\frac{2u}{1-5(u^2)} du = \int\frac{1}{x} dx \\
-\frac{1}{5}\ln|1-5(u^2)| = \ln{x}+C \\
-\frac{1}{5}\ln|1-5(\frac{y}{x})^2| = \ln|x|+C
$$
