To solve the functional equation $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ given by

$$f(xf(x + y)) = yf(x) + 1$$

for all $x, y \in \mathbb{R}^{+}$, we aim to determine a function $f$ that satisfies this relationship.

### Step 1: Assume a functional form

Since $f(xf(x + y)) = yf(x) + 1$, first consider a simple form for $f$, such as $f(x) = \frac{1}{x}$. We need to verify if this candidate satisfies the functional equation.

### Step 2: Verification

Substitute $f(x) = \frac{1}{x}$ into the left-hand side of the given equation:

$$f\left(xf(x + y)\right) = f\left(x \cdot \frac{1}{x+y}\right)$$

This simplifies to:

$$f\left(\frac{x}{x+y}\right) = \frac{x+y}{x}$$

because $f\left(\frac{x}{x+y}\right) = \frac{x+y}{x}$.

Now, consider the right-hand side:

$$yf(x) + 1 = y\left(\frac{1}{x}\right) + 1 = \frac{y}{x} + 1$$

This aligns with the left-hand side since:

$$f\left(\frac{x}{x+y}\right) = \frac{x+y}{x} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}$$

Thus, $f(x) = \frac{1}{x}$ satisfies the functional equation.

### Step 3: Conclusion

The function $f(x) = \frac{1}{x}$ is consistent with the functional equation provided. Therefore, the solution to the problem is:

$$\boxed{f(x) = \frac{1}{x}}$$

This solution meets the criteria for all $x, y \in \mathbb{R}^{+}$ and satisfies the given functional equation throughout the domain of $f$.