3. Let $x$, $y$, $z$ be real numbers, and
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$

Then $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=$
3.1 and -3.

Let $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=k$. Then
$$
\begin{array}{l}
\frac{x}{y+z}=k-\frac{y}{z+x}-\frac{z}{x+y}, \\
\frac{y}{z+x}=k-\frac{z}{x+y}-\frac{x}{y+z}, \\
\frac{z}{x+y}=k-\frac{x}{y+z}-\frac{y}{z+x} .
\end{array}
$$

Substituting the above three equations into
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0
$$

we get $x\left(k-\frac{y}{z+x}-\frac{z}{x+y}\right)+y\left(k-\frac{z}{x+y}-\frac{x}{y+z}\right)+$
$$
z\left(k-\frac{x}{y+z}-\frac{y}{z+x}\right)=0 \text {. }
$$

Simplifying, we obtain $(x+y+z)(k-1)=0$.
If $x+y+z \neq 0$, then $k=1$.
If $x+y+z=0$, then $k=-3$.