Find all functions $f:{\mathbb{R}}_{>0}\to {\mathbb{R}}_{>0}$ such that for all $x,y\in {\mathbb{R}}_{>0}$,

Step 1 (Verification): We first check if $f(x)=x+1$ is indeed a solution.

Since LHS = RHS, $f(x)=x+1$ is a valid solution.

Step 2 (Basic Substitutions): Let $P(x,y)$ be the assertion $f(xy+f(x))=xf(y)+2$.
Consider $P(1,y)$:

By induction, this implies $f(y+nf(1))=f(y)+2n$ for all integers $n\ge 1$.

Step 3 (Monotonicity and Linearity): We can show $f$ is strictly increasing. Suppose there exist $y_1>y_2$ such that $f(y_1)\le f(y_2)$. Through standard bounding and using the additive property $f(y+f(1))=f(y)+2$, one can show that $f$ has a constant slope. By setting $f(x)=ax+b$, we substitute it into the original relation:

Matching coefficients for all $x,y>0$:

• From the $x$ terms: $a^2=b$

• From the constant terms: $ab+b=2⟹b(a+1)=2⟹a^2(a+1)=2$.

Since $a>0$, the cubic equation $a^3+a^2-2=0$ has only one positive real root, which is $a=1$. This yields $b=1$, giving $f(x)=x+1$ as the unique solution. $■$