37th Iranian Mathematical Olympiad

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

Further, putting $y^2$, instead of $y$, in the original equation, one can find that $$af(x{y}^2)+bf\left(\frac{x}{y^2}\right)=cf(x)+g(y^2)$$ for all sufficiently large $y$. Multiplying both sides by $b$, then subtracting, yielding to $$(a^2-ab)f(x{y}^2)=(c^2-2ab-ac)f(x)+T(y)$$ For some function $T(y)$. Putting $y^2$, instead of $y$, we find that $$(a^2-ab)f(x{y}^4)=(c^2-2ab-ac)f(x)+T(y^2).$$ Thus, $$(a^2-ab)(f(x{y}^4)-f(x{y}^2))=T(y^2)-T(y).$$ Hence, if $a\ne b$, we find that $$f(x{y}^4)-f(x{y}^2)=\frac{T(y^2)-T(y)}{a^2-ab}=S(y^2)$$ for some function $S(x)$. That is, $f(x{y}^4)-f(x{y}^2)=S\left(\frac{x{y}^4}{x{y}^2}\right)=S(y^2)$. Therefore, $$\underset{A(y)}{\underbrace{f(x{y}^2)-f(x)}}-\left(\underset{B(y)}{\underbrace{f(x)-f\left(\frac{x}{y^2}\right)}}\right)=R(y).$$ That is, $$f(x{y}^2)+f\left(\frac{x}{y^2}\right)=2f(x)+R(y).$$ 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. ■