Answer
$f(x,y)$ is homogeneous of degree zero.
$$f(v)=\frac{ \sin \frac{1}{v}- v\cos v}{ v}$$
Work Step by Step
Given
$$f(x, y)=\frac{ x \sin\frac{x}{y}- y\cos\frac{y}{x}}{ y}$$
Let
\begin{aligned}
f (t x, ty)&=\frac{ t x \sin\frac{tx}{ty}- ty\cos\frac{ty}{tx}}{ ty}\\
&=\frac{ t x \sin\frac{x}{y}- ty\cos\frac{y}{x}}{ ty}\\
&=\frac{ t( x \sin\frac{x}{y}- y\cos\frac{y}{x})}{ ty}\\
&=\frac{ x \sin\frac{x}{y}- y\cos\frac{y}{x}}{ y}\\
&=f(x,t)
\end{aligned}
Thus, $f(x,y)$ is homogeneous of degree zero.
Transforming the given function
\begin{aligned}f(x, y)&=\frac{ \frac{x}{x} \sin\frac{x}{y}- \frac{y}{x}\cos\frac{y}{x}}{ \frac{y}{x}}\\
&=\frac{ \sin\frac{x}{y}- \frac{y}{x}\cos\frac{y}{x}}{ \frac{y}{x}}\\
\\
\text{Put} \ \ \ v=\frac{y}{x} \ \ \ \text{so, we get}\\
f(x,y)&=\frac{ \sin \frac{1}{v}- v\cos v}{ v}\\
\end{aligned}
