32 Let $\mathbf{R}^{+}$ be the set of all real numbers excluding 0, find all functions $f: \mathbf{R}^{*} \rightarrow \mathbf{R}^{*}$ such that $f(x)+f(y)=f(x y f(x+y))$, where $x$, $y \in \mathbf{R}^{*}$ and $x+y \neq 0$.
We prove that if $f(x)=x_{0}$, then $x=\frac{1}{x_{0}}\left(x, x_{0} \neq 0\right)$.
Fact 1: Otherwise, $f\left(\frac{1}{x_{0}}+x-\frac{1}{x_{0}}\right)=x_{0}$. In the original equation, let $y=x-$
$$
\begin{array}{l}
1 . x \neq \frac{1}{x_{0}} \text { has } f\left(\frac{1}{x_{0}}\right)+f\left(x-\frac{1}{x_{0}}\right)=f\left(\left(x-\frac{1}{x_{0}}\right) \frac{1}{x_{0}} f(x)\right)= \\
f\left(x-\frac{1}{x_{0}}\right) .
\end{array}
$$

Thus, $f\binom{1}{x_{0}}=0$. Contradiction!
Therefore, it must be that $x=\frac{1}{x_{0}}$. This shows that for any $x \in \mathbf{R}^{*}, f(x)=\frac{1}{x}$. Upon verification, it is easy to see that $f(x)=\frac{1}{x}$ satisfies the condition. Hence, $f(x)=\frac{1}{x}$.