2010 Indonesia TST stage 2 test 1 p1 wrote: Find all functions $f : \mathbb R \to \mathbb R$ that satisfies $$xf(y) - yf(x)= f\left(\frac{y}{x}\right)$$ for all $x, y\neq 0 \in \mathbb R$. $P(\frac{1}{x},y)\implies \frac{f(y)}{x}-yf\left(\frac{1}{x}\right)=f(xy)$ $P(1,y)\implies f(y)-yf(1)=f(y)\implies f(1)=0$ $P(x,1)\implies xf(1)-f(x)=f\left(\frac{1}{x}\right)\implies f(x)+f\left(\frac{1}{x}\right)=0$ $\frac{f(y)}{x}+yf(x)=f(xy)$ $P(x,\frac{1}{y})\implies xf\left(\frac{1}{y}\right) - \frac{f(x)}{y}= f\left(\frac{1}{xy}\right)$ $-xf(y) - \frac{f(x)}{y}= -f(xy)\implies xf(y)+\frac{f(x)}{y}=f(xy)$ $\frac{f(y)}{x}+yf(x)=f(xy)=xf(y)+\frac{f(x)}{y}\implies f(y)\frac{x^2-1}{x}=f(x)\frac{y^2-1}{y}\implies f(x)=\frac{x^2-1}{x}c$, where $c$ is just a constant by fixing $y$. Plugging this back into our original equation, we have $$x\frac{y^2-1}{y}c - y\frac{x^2-1}{x}c=\frac{(\frac{y}{x})^2-1}{\frac{y}{x}}c$$ $$x\frac{y^2-1}{y} - y\frac{x^2-1}{x}=\frac{((\frac{y}{x})^2-1)x}{y}$$ $$x\frac{y^2-1}{y} - y\frac{x^2-1}{x}=\frac{\frac{y^2}{x}-x}{y}$$ $$x^2(y^2-1) - y^2(x^2-1)=y^2-x^2$$ This is true, note also that $f(1)=\frac{1^2-1}{1}=0$, therefore $f(x)=\frac{x^2-1}{x}c\,\forall x\neq 0\in \mathbb R$. Answer. $\boxed{f(x)=\frac{x^2-1}{x}c\,\forall x\neq 0\in \mathbb R\text{, where }c\text{ is just an arbitrary constant}}$.

parmenides51 wrote: Find all functions $f : R \to R$ that satisfies $$xf(y) - yf(x)= f\left(\frac{y}{x}\right)$$ for all $x, y \in R$. Let $P(x,y)$ be the assertion $xf(y)-yf(x)=f(\frac yx)$ Let $a=\frac{2f(2)}3$ and $b=f(0)$ $P(1,1)$ $\implies$ $f(1)=0$ and so $P(x,1)$ $\implies$ $f(\frac 1x)=-f(x)$ $P(\frac 1x,2)$ $\implies$ $\frac {3a}{2x}+2f(x)=f(2x)$ $P(\frac 12,x)$ $\implies$ $\frac 12f(x)+\frac {3a}2x=f(2x)$ Subtracting, we get $\boxed{f(0)=b\text{ and }f(x)=a(x-\frac 1x)\quad\forall x\ne 0}$, which indeed fits, whatever are $a,b\in\mathbb R$

Nice FE The only solution is of the form $f(y)=\frac{y^2-1}{y}c$, where $c$ is an arbitrary real number. Let $P(x,y)$ be the given assertion. From $P(1,y)$ we get that $f(1)=0$, from this fact we can easily get from $P(x,1)$ that $-f(x)=f\left(\frac{1}{x}\right)$. This implies that $f(x)+f\left(\frac{1}{x}\right)=0$, plugging $x=-1$ into what we got we get that $f(-1)=0$. From $P(x,-1)$ we get that $f(x)=f\left(-\frac{1}{x}\right)=-f\left(\frac{1}{x}\right)$. This implies that $f$ is an odd function. Adding $P(x,y)$ and $P\left(\frac{1}{x},y\right)$ we have that: $$\left(x+\frac{1}{x}\right)f(y)=f\left(\frac{y}{x}\right)+f(xy)$$ Since we have that $P(x,y)=P\left(\frac{1}{x},\frac{1}{y}\right)$, we get that: $$\frac{y^2-1}{y}f(x)=\frac{x^2-1}{x}f(y)$$ now set $x=2$ now we have that: $$f(y)=\frac{y^2-1}{y}\frac{2f(2)}{3}$$ This implies that $f(y)=\frac{y^2-1}{y}c$, plugging this in we get that this is the solution.