(1) Prove that $f(1) = 1$.
Put $y = 1$ in \textcircled{1}, and write $f(1) = a$. Then
$$f(x) + a + 2x f(x) = \frac{f(x)}{f(x+1)}$$
Thus
$$f(x+1) = \frac{f(x)}{(1 + 2x)f(x) + a} \qquad \textcircled{2}$$
Hence
$$f(2) = \frac{a}{4a} = \frac{1}{4},$$
$$f(3) = \frac{1}{\frac{1}{4} + a} = \frac{1}{5 + 4a},$$
$$f(4) = \frac{1}{7 + 5a + 4a^2}.$$
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On the other hand, we put $x = y = 2$ in \textcircled{1}, then
$$2f(2) + 8f(4) = \frac{f(4)}{f(4)} = 1.$$
By \textcircled{2}, we have
$$\frac{1}{2} + \frac{8}{7 + 5a + 4a^2} = 1.$$
Solving the equation, we have $a = 1$, i.e. $f(1) = 1$.

(2) Prove that
$$f(x+n) = \frac{f(x)}{(n^2 + 2nx)f(x) + 1}, \quad n = 1, 2, \dots \qquad \textcircled{3}$$
Firstly, according to \textcircled{2} we know that \textcircled{3} is true for $n = 1$. Now suppose \textcircled{3} is true for $n = k$. Then
$$\begin{aligned} f(x + k + 1) &= \frac{f(x + k)}{(1 + 2(x + k))f(x + k) + 1} \\ &= \left( \frac{f(x)}{(k^2 + 2kx)f(x) + 1} \right) / \left( \frac{(1 + 2(x + k))f(x)}{(k^2 + 2kx)f(x) + 1} + 1 \right) \\ &= \frac{f(x)}{((k + 1)^2 + 2(k + 1)x)f(x) + 1}. \end{aligned}$$
Hence the result follows by induction.
From \textcircled{3} we have
$$f(n + 1) = \frac{f(1)}{(n^2 + 2n)f(1) + 1} = \frac{1}{(n+1)^2},$$
$$\text{so } f(n) = \frac{1}{n^2}, \quad n = 1, 2, \dots.$$

(3) Prove that
$$f\left(\frac{1}{n}\right) = n^2 = \frac{1}{\left(\frac{1}{n}\right)^2}, \quad n = 1, 2, \dots \qquad \textcircled{4}$$
In fact, by letting $x = \frac{1}{n}$ in \textcircled{3}, we have
$$f\left(n+\frac{1}{n}\right) = \frac{f\left(\frac{1}{n}\right)}{\left(n^2+2\right)f\left(\frac{1}{n}\right)+1},$$
and by setting $y = \frac{1}{x}$ in \textcircled{1}, we have
$$f(x) + f\left(\frac{1}{x}\right) + 2 = \frac{1}{f\left(x + \frac{1}{x}\right)}.$$
So
$$f(n) + f\left(\frac{1}{n}\right) + 2 = \frac{1}{f\left(n + \frac{1}{n}\right)} = n^2 + 2 + \frac{1}{f\left(\frac{1}{n}\right)}.$$
Consequently, $f(n) = \frac{1}{n^2}$ implies $f(\frac{1}{n}) = n^2$.

(4) Prove that if $q = \frac{n}{m}$, $\text{gcd}(m, n) = 1$, $m, n \in \mathbb{N}^*$, then $f(q) = \frac{1}{q^2}$.
For $m, n \in \mathbb{N}^*$, $\text{gcd}(m, n) = 1$, put $x = n$, $y = \frac{1}{m}$ in \textcircled{1}, we have
$$f\left(\frac{1}{m}\right) + f(n) + \frac{2n}{m}f\left(\frac{n}{m}\right) = \frac{f\left(\frac{n}{m}\right)}{f\left(n+\frac{1}{m}\right)}.$$
Put $x = \frac{1}{m}$ in \textcircled{3}, we get
$$f\left(n+\frac{1}{m}\right) = \frac{f\left(\frac{1}{m}\right)}{\left(n^2+\frac{2n}{m}\right)f\left(\frac{1}{m}\right)+1} = \frac{1}{n^2+\frac{2n}{m}+\frac{1}{m^2}}.$$
So
$$\frac{1}{n^2} + m^2 + \frac{2n}{m} f\left(\frac{n}{m}\right) = \left(n + \frac{1}{m}\right)^2 f\left(\frac{n}{m}\right).$$
Now we have
$$f(q) = f\left(\frac{n}{m}\right) = \frac{\frac{1}{n^2} + m^2}{n^2 + \frac{1}{m^2}} = \left(\frac{m}{n}\right)^2 = \frac{1}{q^2}.$$
Finally, it is easy to verify that $f(x) = \frac{1}{x^2}$ satisfies the condition. So $f(x) = \frac{1}{x^2}$ is the answer.