What relationship exists between $A$ and $p$ if

$$
A=\frac{x^{2}-3 y^{2}}{3 x^{2}+y^{2}}
$$

and

$$
\frac{p x y}{x^{2}-(2+p) x y+2 p y^{2}}-\frac{y}{x-2 y}=\frac{1}{2} ?
$$
The case where $x=y=0$ must obviously be excluded, but it is also not possible for $y=0, x \neq 0$, because then (3) would not hold. If $x=0, y \neq 0$, then $A=-3$, and (2) would turn into the identity $\frac{1}{2}=\frac{1}{2}$, and thus $p$ could be anything, meaning there is no relationship between $A$ and $p$. Let us assume from now on that neither $x$ nor $y$ is 0. We also assume that none of the denominators in (3) is 0. The first denominator can be factored as: $(x-2 y)(x-p y)$, so $x \neq 2 y, x \neq p y$. Simplifying the fractions by $y^{2}$ and $y$ respectively, the right-hand side of (2) and the left-hand side of (3) can be expressed in terms of the variable $x / y=t$:

$$
A 7 \frac{t^{3}-3}{3 t^{2}+1}, \quad\left(2^{\prime}\right) \quad \frac{p t}{(t-2)(t-p)}-\frac{1}{t-2}=\frac{1}{2}
$$

Multiplying (3') by $2(t-2)(t-p)$ and rearranging, we get

$$
t^{2}-3 p t=t(t-3 p)=0
$$

Thus, since $t \neq 0$, the assumption (3) states that the value of $t$ can only be $3 p$. Therefore, the sought relationship from (2') is

$$
A=\frac{9 p^{2}-3}{27 p^{2}+1}
$$