Putting $xy$, $\frac{x}{y}$ instead of $x$, we find that
$$a f(xy^2) + b f(y) = c f(xy) + g(y)$$
and,
$$a f(x) + b f\left(\frac{x}{y^2}\right) = c f\left(\frac{x}{y}\right) + g(y).$$
Multiplying the first equation by $a$ and the second one by $b$, then, adding and using the original equation, we shall find that
$$a^2 f(xy^2) + b^2 f\left(\frac{x}{y^2}\right) = (c^2 - 2ab)f(x) + (a + b + c)g(y).$$

Further, putting $y^2$, instead of $y$, in the original equation, one can find that
$$a f(xy^2) + b f\left(\frac{x}{y^2}\right) = c f(x) + g(y^2)$$
for all sufficiently large $y$. Multiplying both sides by $b$, then subtracting, yielding to
$$(a^2 - ab) f(xy^2) = (c^2 - 2ab - ac) f(x) + T(y)$$
For some function $T(y)$. Putting $y^2$, instead of $y$, we find that
$$(a^2 - ab) f(xy^4) = (c^2 - 2ab - ac) f(x) + T(y^2).$$
Thus,
$$(a^2 - ab) (f(xy^4) - f(xy^2)) = T(y^2) - T(y).$$
Hence, if $a \neq b$, we find that
$$f(xy^4) - f(xy^2) = \frac{T(y^2) - T(y)}{a^2 - ab} = S(y^2)$$
for some function $S(x)$. That is, $f(xy^4) - f(xy^2) = S\left(\frac{xy^4}{xy^2}\right) = S(y^2)$.
Therefore,
$$\underbrace{f(xy^2) - f(x)}_{A(y)} - \left( \underbrace{f(x) - f\left(\frac{x}{y^2}\right)}_{B(y)} \right) = R(y).$$
That is,
$$f(xy^2) + f\left(\frac{x}{y^2}\right) = 2f(x) + R(y).$$
Putting $\sqrt{y}$ instead of $y$ we will eventually arrive at the desired conclusion. The remaining cases for $c = 2a$ and $c = -a$ are the same. ■