3. Find the positive rational number pairs $(x, y)$ that satisfy $y x^{y}=y+1$.

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3. Let $y=\frac{p}{q}\left((p, q)=1, p, q \in \mathbf{Z}_{+}\right)$.

Then the original equation can be rewritten as
$$
\begin{array}{l}
x^{\frac{p}{q}}=\frac{y+1}{y}=\frac{\frac{p}{q}+1}{\frac{p}{q}}=\frac{p+q}{p} \\
\Rightarrow x=\sqrt[p]{\frac{(p+q)^{q}}{p^{q}}} \in \mathbf{Q} .
\end{array}
$$

Since $(p+q, q)=1$, then $(p+q)^{q}$ and $p^{q}$ are both powers of $p$. Without loss of generality, assume
$$
p+q=a^{p}, p=b^{p}\left(a, b \in \mathbf{Z}_{+}, a>1\right) .
$$

If $b \geqslant 2$, then $p=b^{p} \geqslant 2^{p}>p$, which is a contradiction.
Thus, $b=1 \Rightarrow p=1, q=a-1$.
$$
\text { Hence }(x, y)=\left(a^{a-1}, \frac{1}{a-1}\right)\left(a \in \mathbf{Z}_{+}, a>1\right) \text {. }
$$