Find all functions $f:\mathbb R \to \mathbb R$ satisfying $xf(y)-yf(x)=f\left( \frac yx\right)$

Find all functions $f:\mathbb R \to \mathbb R$ that satisfy the following equation: $$xf(y)-yf(x)=f\left( \frac yx\right).$$

My work so far

1. If $x=1$ then $f(1)=0$

2. If $y=1$ then $f\left(\frac1x\right)=-f(x)$

3. If $y=\frac1x$ then $f\left(x^2\right)=\left(x+\frac1x\right)f(x)$

Solution 1:

This may not be the most efficient approach, but it works. I'll try to trim it down later.

You have already found that $f(x)=-f(\tfrac{1}{x})$, so then plugging in $x=-1$ shows that $f(-1)=0$. Next taking $y=-x$ in the original equation yields $$xf(-x)+xf(x)=f(-1)=0,$$ which shows that $f(-x)=-f(x)$ for all $x\in\Bbb{R}$. It follows that for all nonzero $x\in\Bbb{R}$ we have $$f(-\tfrac{1}{x})=f(x).$$ Now let $x,y\in\Bbb{R}$ be nonzero. Then writing out the original equation for $-\tfrac{1}{x}$ and $-\tfrac{1}{y}$ yields $$-\tfrac{1}{x}f(-\tfrac{1}{y})+\tfrac{1}{y}f(-\tfrac{1}{x})=f(\tfrac{x}{y})=-f(\tfrac{y}{x})=-xf(y)+yf(x).$$ But on the left hand side we can use the fact that $f(-\tfrac{1}{x})=f(x)$ to get $$-\tfrac{1}{x}f(y)+\tfrac{1}{y}f(x)=-xf(y)+yf(x).$$ Rearranging terms shows that for all nonzero $x,y\in\Bbb{R}$ with $x,y\neq\pm1$ we have $$\frac{f(x)}{x-\tfrac{1}{x}}=\frac{f(y)}{y-\tfrac{1}{y}},$$ from which it follows that there is some constant $c\in\Bbb{R}$ such that $$f(x)=c(x-\tfrac1x),$$ for all nonzero $x\in\Bbb{R}$, as we already saw that $f(\pm1)=0$. Plugging in $y=0$ shows that also $f(0)=0$, and so every such function $f$ is (for some $c\in\Bbb{R}$) of the form $$f(x)=\left\{\begin{array}{ll}0&\text{ if }x=0\\c(x-\tfrac{1}{x})&\text{ otherwise }\end{array}\right.$$

Solution 2:

$$xf(y)-yf(x)=f\left( \frac yx\right) \\ \iff xf(y)-f(\frac{y}{x})-yf(x) = 0 \tag{1}$$

Substitute $x \to y/x$, $y \to y$:

$$\frac{y}{x}f(y) - yf( \frac{y}{x}) = f(x) \\ \iff \frac{y}{x}f(y) - yf( \frac{y}{x})-f(x) = 0 \tag{2}$$

Consider $(2)-(1)\times y$:

$$(\frac{y}{x}-xy)f(y)+(y^2-1)f(x)=0 \tag{3}$$

Let $y = 0$, As M. Vinay said, $f(0) = 0$

$0\times f(0) + (0-1)f(x)=0 \implies f(x) = 0$ for all $x \in \mathbb{R}$

EDIT: My setting value is wrong(i.e. after formulate (3) , all what I did is wrong ) so I want to set them right.

Set $y = 0.5$ in (3), $$(\frac{1}{2x}-\frac{x}{2}) f(1/2) - 3/4 f(x) = 0 \\ \iff f(x) =(\frac{2}{3x}-\frac{2x}{3}) f(1/2) $$

More precisely, $ f(x) = \frac{2}{3} f(1/2) \times(\frac{1}{x}-x)$

Let $\frac{2}{3} f(1/2) = c$ so$ f(x) = c(\frac{1}{x}-x)$

Since, c is an arbitrary constant, it includes my special solution $f(x) = 0$.