3. Let the even function $f(x)$ satisfy: $f(1)=2$, and when $x y$ $\neq 0$, $f\left(\sqrt{x^{2}+y^{2}}\right)=\frac{f(x) f(y)}{f(x)+f(y)}$. Then $f(5)$ $=$
3. $\frac{2}{25}$.

From the condition, we have $\frac{1}{f\left(\sqrt{x^{2}+y^{2}}\right)}=\frac{1}{f(x)}+\frac{1}{f(y)}$.
Since $f(-x)=f(x)$, we only need to consider the case where $x, y \in \mathbf{R}_{+}$.
Let $x=\sqrt{u}, y=\sqrt{v}$. Then
$\frac{1}{f(\sqrt{u+v})}=\frac{1}{f(\sqrt{u})}+\frac{1}{f(\sqrt{v})}$.
Let $g(t)=\left\{\begin{array}{cc}\frac{1}{f(\sqrt{t})}, & f(\sqrt{t}) \neq 0 ; \\ 0, & f(\sqrt{t})=0 .\end{array}\right.$
Then it becomes $g(u+v)=g(u)+g(v)$.
From equation (1), we know $g(2 u)=g(u+u)=2 g(u)$,
$g(3 u)=g(u+2 u)=g(u)+2 g(u)$
$=3 g(u)$.
By induction, we get $g(k u)=k g(u)$.
Let $k=25, u=1$, we get $g(25)=25 g(1)$, that is
$$
\frac{1}{f(\sqrt{25})}=\frac{25}{f(\sqrt{1})} \Rightarrow f(5)=\frac{2}{25} .
$$