Let $P(x, y)$ be the assertion $f\left(\frac{y}{f(x+1)}\right) + f\left(\frac{x+1}{x f(y)}\right) = f(y)$.

Rewriting the equation, we get $f\left(\frac{y}{f(x)}\right) + f\left(\frac{x}{(x-1)f(y)}\right) = f(y)$ which is valid for each $x > 1,\ y \in \mathbb{R}^+$.

If there exists $a \in \mathbb{R}^+$ for which $f(a) > \frac{1}{a}$, we get
$$P\left(\frac{a f(a)}{a f(a)-1}, a\right) \Rightarrow f\left(\frac{a}{f\left(\frac{a f(a)}{a f(a)-1}\right)}\right) = 0,$$
which is a contradiction. Therefore, for each $x \in \mathbb{R}^+$, $f(x) \le \frac{1}{x}$.

Now, we can write
$$P(x > 1, y) \Rightarrow f(y) = f\left(\frac{y}{f(x)}\right) + f\left(\frac{x}{(x-1)f(y)}\right) \le \frac{f(x)}{y} + \left(1 - \frac{1}{x}\right)f(y) \Rightarrow y f(y) \le x f(x).$$
So for all $x, y > 1 \in \mathbb{R}^+$, $x f(x) = y f(y)$. Hence for each $x > 1 \in \mathbb{R}^+$, $f(x) = \frac{C}{x}$ where $C$ is a constant number. If we choose real numbers $x, y$ greater than 1 such that $y$ is also greater than $C$, substituting these values for $x$ and $y$ in the equation shows that $C = 1$. Therefore, $\forall x > 1 \in \mathbb{R}^+$, $f(x) = \frac{1}{x}$.

Since $f(1) \le 1$, we have
$$\begin{aligned} P(x > 1, 1) &\Rightarrow f(1) = f\left(\frac{1}{f(x)}\right) + f\left(\frac{x}{(x-1)f(1)}\right) = f(x) + \frac{x-1}{x}f(1) \\ &\Rightarrow \frac{1}{x}f(1) = \frac{1}{x} \Rightarrow f(1) = 1. \end{aligned}$$
Now, for $\frac{1}{2} \le x < 1$, we can write
$$\begin{aligned} P(2, x) &\Rightarrow f\left(\frac{x}{f(2)}\right) + f\left(\frac{2}{f(x)}\right) = f(x) \Rightarrow f(2x) + f\left(\frac{2}{f(x)}\right) = f(x) \\ &\Rightarrow \frac{1}{2x} + \frac{f(x)}{2} = f(x) \Rightarrow f(x) = \frac{1}{x}. \end{aligned}$$
In a similar way, by using induction on $n$ one can prove that for each positive real $x$ in the interval $\frac{1}{2^n} \le x < \frac{1}{2^{n-1}}$, $f(x) = \frac{1}{x}$. Therefore, for each $x \in \mathbb{R}^+$, $f(x) = \frac{1}{x}$. It's easy to verify that this solution is indeed an answer to the functional equation.