Putting $x = y = 1$ in (2) yields $f(1) = f(1)^2$, that is $f(1) \in \{0, 1\}$.

Case 1: $f(1) = 0$
(a) If $k = 0$, (2) becomes for $x = y = t$, $t \neq 0$: $f(1) = t^2 f(t)^2$, whence $f(t) = 0$, $t \neq 0$. Let $k \neq 0$. Then $y = 1$ and $x = t$ in (2) yields $f(t^k) = t f(t) f(1) = 0$. Letting $x = y$ in (2) finally leads to $f((x^2)^k) = x^2 f(x)^2$, that is $0 = x^2 f(x)^2$, whence $f(x) = 0$, $x \neq 0$.

Case 2: $f(1) = 1$
Letting $x = t$ and $y = 1/t$, $t \neq 0$, in (2) shows $f(1) = f(t) \cdot f(1/t)$. Therefore, $f(t) \neq 0$, $t \neq 0$. Putting $x = y = -1$ in (2) yields $f((-1)^{2k}) = (-1)^2 f(-1)^2$, that is $f(-1)^2 = 1$. Thus $f(-1) \in \{-1, 1\}$.
(a) Let $k$ be odd. Then $x = 1$ and $y = -1$ in (2) leads to
$$f((-1)^k) = -f(-1) \qquad (3)$$
that is $f(-1) = -f(-1)$. Therefore, we get the contradiction $f(-1) = 0$.
(b) For $k$ even we get from (3): $f(-1) = -1$.

i) If $k = 0$, (2) yields for $y = 1$ and $x = t$, $t \neq 0$: $f(1) = t f(t)$ that is $f(t) = 1/t$.

ii) For $k \neq 0$ we get from (2) for $x = 1$, $y = t$, $t \neq 0$:
$$f(t^k) = t f(t). \qquad (4)$$
This and (2) imply
$$xyf(x)f(y) = f(x^k y^k) = f((xy)^k) = xyf(xy).$$
Therefore,
$$f(xy) = f(x)f(y) \qquad (5)$$
Putting $x = t$ and $y = t^{k-1}$, $t \neq 0$, yields in view of (4) and (5):
$$tf(t) = f(t^k) = f(t \cdot t^{k-1}) = f(t)f(t^{k-1}).$$
Therefore, $f(t^{k-1}) = t$, $t \neq 0$. As $k$ is even all real numbers $x$, $x \neq 0$, have a unique representation $x = t^{k-1}$, $t \neq 0$. Thus, finally $f(x) = x^{1/(k-1)} = t^{k-1/2}$.

It is easily checked that all functions obtained in the course of our solution satisfy (2).

Summary: All solutions to (2) are given by
* $f(x) = 0$, $x \in \mathbb{R}$, for all integers $k$ and additionally
* $f(x) = \begin{cases} x^{1/(k-1)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$, if $k$ is an even integer.