Answer
Homogeneous with a degree of 1
Work Step by Step
Recall that the definition of a homogeneous function is a function that satisfies the following condition
$f(tx,ty)=t^nf(x,y)$ where $n$ is the degree.
If $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$
all we have to do is plug in the point $(tx,ty)$
$f(xt,yt)=\frac{(tx)(ty)}{\sqrt{(tx)^2+(ty)^2}}$
$=\frac{t^2xy}{\sqrt{t^2x^2+t^2y^2}}=\frac{t^2xy}{\sqrt{t^2(x^2+y^2)}}=\frac{t^2xy}{t\sqrt{x^2+y^2}}=\frac{t^1xy}{\sqrt{x^2+y^2}}=tf(x,y)$
So the function is homogeneous with a degree of 1
