Solution:
Putting $x=0, y=0$, we get $f(0)=f(0)^3$ so that $f(0)=0, 1$ or $-1$.
If $f(0)=0$, then taking $y=0$ in the given equation, we obtain $f(x)=f(x) f(0)^2=0$ for all $x$.

Suppose $f(0)=1$. Taking $y=-x$, we obtain
$$1=f(0)=f(x-x)=f(x) f(-x) f\left(-x^2\right)$$
This shows that $f(x) \neq 0$ for any $x \in \mathbf{R}$.

Taking $x=1, y=x-1$, we obtain
$$f(x)=f(1) f(x-1)^2=f(1)[f(x) f(-x) f(-x)]^2$$
Using $f(x) \neq 0$, we conclude that $1=k f(x)(f(-x))^2$, where $k=f(1)(f(-1))^2$.
Changing $x$ to $-x$ here, we also infer that $1=k f(-x)(f(x))^2$.
Comparing these expressions we see that $f(-x)=f(x)$.
It follows that $1=k f(x)^3$.
Thus $f(x)$ is constant for all $x$.
Since $f(0)=1$, we conclude that $f(x)=1$ for all real $x$.

If $f(0)=-1$, a similar analysis shows that $f(x)=-1$ for all $x \in \mathbf{R}$.

We can verify that each of these functions satisfies the given functional equation. Thus there are three solutions, all of them being constant functions.