Answer
$f(x,y)$ is homogeneous of degree zero.
$$f(x,y)=\sqrt{ 1+ v^2}$$
Work Step by Step
Given
$$f(x, y)=\frac{\sqrt{x^2+y^2}}{x} $$
Let
\begin{aligned}
f (t x, ty)&=\frac{\sqrt{(tx)^2+(ty)^2}}{tx}\\
&=\frac{t(3x+10y)}{6 ty}\\
&=\frac{t\sqrt{x^2+y^2}}{tx}\\
&=\frac{\sqrt{x^2+y^2}}{x}\\
&=f(x,t)
\end{aligned}
Thus, $f(x,y)$ is homogeneous of degree zero.
Transforming the given function
\begin{aligned}f(x, y)&=\frac{\sqrt{ \frac{x^2}{x^2}+ \frac{y^2}{x^2}}}{\frac{x}{x}}\\
&=\sqrt{ 1+ \frac{y^2}{x^2}}
\\
\text{Put} \ \ \ v=\frac{y}{x} \ \ \ \text{so, we get}\\
f(x,y)&=\sqrt{ 1+ v^2}
\end{aligned}
