Let $\mathbb{R}^{+}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that, for all positive real numbers $x$ and $y$, 
\[ f(x+f(x y))+y=f(x) f(y)+1 \]
A straightforward check shows that \( f(x) = x + 1 \) satisfies (*). We divide the proof of the converse statement into a sequence of steps. Step 1: \( f \) is injective. Put \( x = 1 \) in (*) and rearrange the terms to get 
\[ y = f(1) f(y) + 1 - f(1 + f(y)) \]
Therefore, if \( f(y_1) = f(y_2) \), then \( y_1 = y_2 \). Step 2: \( f \) is (strictly) monotone increasing. For any fixed \( y \in \mathbb{R}^+ \), the function 
\[ g(x) := f(x + f(xy)) = f(x) f(y) + 1 - y \]
is injective by Step 1. Therefore, \( x_1 + f(x_1 y) \neq x_2 + f(x_2 y) \) for all \( y, x_1, x_2 \in \mathbb{R}^+ \) with \( x_1 \neq x_2 \). Plugging in \( z_i = x_i y \), we arrive at 
\[ \frac{z_1 - z_2}{y} \neq f(z_2) - f(z_1), \quad \text{or} \quad \frac{1}{y} \neq \frac{f(z_2) - f(z_1)}{z_1 - z_2} \]
for all \( y, z_1, z_2 \in \mathbb{R}^+ \) with \( z_1 \neq z_2 \). This means that the right-hand side of the rightmost relation is always non-positive, i.e., \( f \) is monotone non-decreasing. Since \( f \) is injective, it is strictly monotone. Step 3: There exist constants \( a \) and \( b \) such that \( f(y) = a y + b \) for all \( y \in \mathbb{R}^+ \). Since \( f \) is monotone and bounded from below by 0, for each \( x_0 \geqslant 0 \), there exists a right limit \( \lim_{x \searrow x_0} f(x) \geqslant 0 \). Put \( p = \lim_{x \searrow 0} f(x) \) and \( q = \lim_{x \searrow p} f(x) \). Fix an arbitrary \( y \) and take the limit of (*) as \( x \searrow 0 \). We have \( f(xy) \searrow p \) and hence \( f(x + f(xy)) \searrow q \); therefore, we obtain 
\[ q + y = p f(y) + 1, \quad \text{or} \quad f(y) = \frac{q + y - 1}{p} \]
(Notice that \( p \neq 0 \), otherwise \( q + y = 1 \) for all \( y \), which is absurd.) The claim is proved. Step 4: \( f(x) = x + 1 \) for all \( x \in \mathbb{R}^+ \). Based on the previous step, write \( f(x) = a x + b \). Putting this relation into (*) we get 
\[ a(x + a x y + b) + b + y = (a x + b)(a y + b) + 1, \]
which can be rewritten as 
\[ (a - a b) x + (1 - a b) y + a b + b - b^2 - 1 = 0 \quad \text{for all} \ x, y \in \mathbb{R}^+ . \]
This identity may hold only if all the coefficients are 0, i.e., 
\[ a - a b = 1 - a b = a b + b - b^2 - 1 = 0 . \]
Hence, \( a = b = 1 \).