Let $f: \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that for all $x, y \in \mathbb{R}^+$, the following functional equation holds:

$$(1 + y f(x))(1 - y f(x+y)) = 1.$$

Our goal is to find all such functions $f$.

### Step 1: Simplify the Functional Equation

Expanding the equation, we have:

$$1 + y f(x) - y f(x+y) - y^2 f(x) f(x+y) = 1.$$

Subtracting 1 from both sides gives:

$$y f(x) - y f(x+y) - y^2 f(x) f(x+y) = 0.$$

We can factor out $y$ from the terms:

$$y (f(x) - f(x+y)) = y^2 f(x) f(x+y).$$

Assuming $y \neq 0$, we divide both sides by $y$:

$$f(x) - f(x+y) = y f(x) f(x+y).$$

### Step 2: Analyze the Equation

This can be rewritten as:

$$f(x) = f(x+y) + y f(x) f(x+y).$$

### Step 3: Assume a Form for $f(x)$

Assume $f(x) = \frac{1}{x + a}$ for some constant $a > 0$. We will verify if this function satisfies the given functional equation.

Substituting this form into the right-hand side of the equation yields:

$$f(x+y) = \frac{1}{x+y+a}.$$

Check:

$$(1 + y \cdot \frac{1}{x+a}) \left( 1 - y \cdot \frac{1}{x+y+a} \right) = 1.$$

Simplify $1 + \frac{y}{x+a}$:

$$1 + \frac{y}{x+a} = \frac{x+a+y}{x+a}.$$

Simplify $1 - \frac{y}{x+y+a}$:

$$1 - \frac{y}{x+y+a} = \frac{x+y+a-y}{x+y+a} = \frac{x+a}{x+y+a}.$$

Substituting both expressions back into the equation:

$$\frac{x+a+y}{x+a} \cdot \frac{x+a}{x+y+a} = 1.$$

The expressions simplify to:

$$\frac{(x+a)(x+a+y)}{(x+a)(x+y+a)} = 1.$$

Thus, the equality holds, verifying our assumption.

Therefore, the required function is:

$$f(x) = \frac{1}{x + a}.$$

The function satisfies the equation for any positive constant $a$, so the solution is:

$$\boxed{f(x) = \frac{1}{x + a}}$$
for any positive constant $a > 0$, which completes the solution.