3. The function $f: R \rightarrow R$, for any real numbers $x, y$, as long as $x+y \neq 0$, then $f(x y)=$ $\frac{f(x)+f(y)}{x+y}$ holds, then the odd-even property of the function $f(x)(x \in \mathbf{R})$ is $(\quad)$.
A. Odd function
B. Even function
C. Both odd and even function
D. Neither odd nor even function
3. $\mathrm{C}$

Since $f\left(x^{2}\right)=\frac{f(x)+f(x)}{x+x}=\frac{f(x)}{x}$,
$$
f\left(x^{2}\right)=f[(-x)(-x)]=\frac{f(-x)}{-x}, x \neq 0,
$$

Therefore, $f(x)=-f(-x)$, where $x \neq 0$.
So $f(0)=f(0 \times 1)=\frac{f(0)+f(1)}{1}$,
So $f(1)=0$, so $f(-1)=-f(1)=0$,
So $f(0)=f(0 \times(-1))=\frac{f(0)+f(-1)}{-1}=-f(0)$.
So $f(0)=0$.
Also, since $f(x)=f(1 \times x)=\frac{f(1)+f(x)}{x+1}=\frac{f(x)}{x+1}(x \neq 0,-1)$,
So $f(x)=0$.
Therefore, $f(x)=0(x \in \mathbf{R})$.