Answer
$f(x,y)$ is homogeneous of degree zero.
$$f(x,y)=\frac{\sqrt{1+4 v^2}-1+v}{1+3v}$$
Work Step by Step
Given
$$f(x, y)=\frac{\sqrt{x^2+4y^2}-x+y}{x+3y} $$
Let
\begin{aligned}
f (t x, ty)&=\frac{\sqrt{(tx)^2+4(ty)^2}-tx+ty}{tx+3ty}\\
&=\frac{t\sqrt{x^2+4y^2}-tx+ty}{tx+3ty}\\
&=\frac{t(\sqrt{x^2+4y^2}-x+y)}{t(x+3y)}\\
&=\frac{\sqrt{x^2+4y^2}-x+y}{x+3y}\\
&=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}+4 \frac{y^2}{x^2}}-\frac{x}{x}+\frac{y}{x}}{\frac{x}{x}+3\frac{y}{x}}\\
&=\frac{\sqrt{1+4 \frac{y^2}{x^2}}-1+\frac{y}{x}}{1+3\frac{y}{x}}\\
\\
\text{Put} \ \ \ v=\frac{y}{x} \ \ \ \text{so, we get}\\
f(x,y)&=\frac{\sqrt{1+4 v^2}-1+v}{1+3v}\\
\end{aligned}
