Example 1. Find the general integral of the equation $y^{\prime}=\frac{x^{2}+3 x y-y^{2}}{3 x^{2}-2 x y}$.
Solution. Formally, it is not obvious that the right-hand side of this equation is a function depending on the ratio $\frac{y}{x}$. To see this, factor out $x^2$ in the numerator and the denominator and cancel this factor. We get

$$
y'=\frac{x^{2}\left(1+3 \frac{y}{x}-\left(\frac{y}{x}\right)^{2}\right)}{x^{2}\left(3-2 \frac{y}{x}\right)}, \quad y'=\frac{1+3 \frac{y}{x}-\left(\frac{y}{x}\right)^{2}}{3-2 \frac{y}{x}}
$$

We have obtained a homogeneous equation, the form of which is denoted in point $1^{\circ}$.

Let $y=t x, y'=t' x+t$ (here $t$ is the new unknown function of $x$). We get the equation $t' x+t=\frac{1+3 t-t^{2}}{3-2 t}$. Transforming, by moving $t$ to the right side: $t' x=\frac{1+3 t-t^{2}-3 t+2 t^{2}}{3-2 t}$, or $\frac{d t}{d x} \cdot x=\frac{1+t^{2}}{3-2 t}$. Separating variables: $\frac{3-2 t}{1+t^{2}} d t=\frac{d x}{x}$. Integrating: $3 \operatorname{arctg} t-\ln \left(1+t^{2}\right)=$ $=\ln |x|+\ln C$, or $3 \operatorname{arctg} t=\ln \left(C|x|\left(1+t^{2}\right)\right)$. Substituting $t=\frac{y}{x}$,
we obtain the desired general integral.

Answer. $3 \operatorname{arctg} \frac{y}{x}=\ln \frac{C\left(x^{2}+y^{2}\right)}{|x|}$.