3. Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that for any $x, y \in \mathbf{R}_{+}$, we have
$$
f\left(\frac{f(y)}{f(x)}+1\right)=f\left(x+\frac{y}{x}+1\right)-f(x) .
$$
3. $f(x)=a x\left(a \in \mathbf{R}_{+}\right)$.

Taking $y=x$, we get
$$
f(x+2)-f(x)=f(2)>0 \text {. }
$$

First, we prove that $f(x)$ is injective.
Take positive real numbers $x_{1} 、 x_{2}$, and $x_{2}-x_{1}>1$.
Take $x=x_{1}, y=x_{1}\left(x_{2}-x_{1}-1\right)$.
Then the left side of the original equation is a positive number, while the right side is $f\left(x_{2}\right)-f\left(x_{1}\right)$. Therefore, $f\left(x_{2}\right)>f\left(x_{1}\right)$.
If $f\left(x_{1}\right)=f\left(x_{2}\right)$, then it must be that $\left|x_{1}-x_{2}\right| \leqslant 1$.
Assume there is a pair of positive real numbers $a 、 b(a \neq b)$,
$$
f(a)=f(b) \text {. }
$$

Taking $x=a$ and $x=b$ respectively, we get for any positive number $y$,
$$
f\left(a+\frac{y}{a}+1\right)=f\left(b+\frac{y}{b}+1\right) .
$$

When $y$ is very large,
$$
\begin{array}{l}
\left|\left(a+\frac{y}{a}+1\right)-\left(b+\frac{y}{b}+1\right)\right| \\
=\left|y\left(\frac{1}{a}-\frac{1}{b}\right)+a-b\right|>1,
\end{array}
$$

which is a contradiction.
Therefore, $f(x)$ is injective.
Taking $x=2$, we get
$$
\begin{array}{l}
f\left(\frac{f(y)}{f(2)}+1\right)=f\left(\frac{y}{2}+3\right)-f(2) \\
=f\left(\frac{y}{2}+1\right) .
\end{array}
$$

Since $f$ is injective, we have
$$
\begin{array}{l}
\frac{f(y)}{f(2)}+1=\frac{y}{2}+1 \\
\Rightarrow f(y)=\frac{f(2)}{2} y .
\end{array}
$$

Thus, $f(x)=a x$, where $a$ is a positive constant.
It is easy to verify that $f(x)=a x$ satisfies the conditions of the problem.