First, prove that $f \le \frac{1}{7}$; when $x = y = z = \frac{1}{3}$, we have $f = \frac{1}{7}$.
Since $f = \sum \frac{x(x+3y-1)}{1+x+3y} = 1 - 2\sum \frac{x}{1+x+3y}$, by Cauchy's inequality
$$\sum \frac{x}{1+x+3y} \ge \frac{\left(\sum x\right)^2}{\sum x(1+x+3y)} = \frac{1}{\sum x(1+x+3y)},$$
and
$$\sum x(1+x+3y) - \sum x(2x+4y+z) = 2 + \sum xy \le \frac{7}{3}.$$
So $\sum \frac{x}{1+x+3y} \ge \frac{3}{7}$, $f \le 1 - 2 \times \frac{3}{7} = \frac{1}{7}$; $f_{max} = \frac{1}{7}$; when $x = y = z = \frac{1}{3}$, we have $f = \frac{1}{7}$.

Second, prove that $f \ge 0$; when $x = 1, y = z = 0$, we have $f = 0$.
In fact, one can see that
$$\begin{aligned} f(x, y, z) &= \frac{x(2y-z)}{1+x+3y} + \frac{y(2z-x)}{1+y+3z} + \frac{z(2x-y)}{1+z+3x} \\ &= xy \left( \frac{2}{1+x+3y} - \frac{1}{1+y+3z} \right) \\ &\quad + xz \left( \frac{2}{1+z+3x} - \frac{1}{1+x+3y} \right) \\ &\quad + yz \left( \frac{2}{1+y+3z} - \frac{1}{1+z+3x} \right) \\ &= \frac{7xyz}{(1+x+3y)(1+y+3z)} \\ &\quad + \frac{7xyz}{(1+z+3x)(1+x+3y)} \\ &\quad + \frac{7xyz}{(1+y+3z)(1+z+3x)} \\ &\ge 0. \end{aligned}$$