11. 3. Find all functions $f(x)$ that satisfy the following conditions: defined and positive for all positive $x$, and for all positive $x, y$ the equation $f\left(x^{y}\right)=f(x)^{f(y)}$ holds.
11.3. $1, x$.

First, $f(x) \equiv 1$ satisfies the given conditions. Now let's find other solutions.
Assume for some $a>0, f(a) \neq 1$. Then from the equation
$$
f(a)^{f(x y)}=f\left(a^{x y}\right)=f\left(a^{x}\right)^{f(y)}=f(a)^{f(x) f(y)}
$$

we get $f(x y)=f(x) f(y)$
for any $x, y>0$. And from the equation
$$
\begin{array}{l}
f(a)^{f(x+y)}=f\left(a^{x+y}\right)=f\left(a^{x}\right) \cdot f\left(a^{y}\right) \\
=(f(a))^{f(x)} \cdot(f(a))^{f(y)}=(f(a))^{f(x)+f(y)}
\end{array}
$$

we get $f(x+y)=f(x)+f(y)$
for any $x, y>0$.
From (1) we have $f(1)=f(1 \cdot 1)=(f(1))^{2}$, i.e., $f(1)=1$. Then from (2) and (1) we have
$$
\begin{array}{l}
f(n)=f(1+\cdots+1)=f(1)+\cdots+f(1)=n, \\
f\left(\frac{m}{n}\right) \cdot n=f\left(\frac{m}{n}\right) \cdot f(n)=f(m)=m .
\end{array}
$$

i.e., for any $m, n \in N, f\left(\frac{m}{n}\right)=\frac{m}{n}$.
Assume for some $x>0$ there is $f(x) \neq x$, for example $f(x)x\right.$ the case is similar). Take $y=\frac{m}{n}$ such that $f(x)f(y)=y .
\end{aligned}
$$