Solution:
Taking $x=y=0$ in (1), we get $z f(0)=f(0)$ for all $z \in \mathbf{R}$. Hence we obtain $f(0)=0$.

Taking $y=0$ in (1), we get
$$f\left(x^{2}\right)=x f(x)$$
Similarly $x=0$ in (1) gives
$$f(y f(z))=z f(y)$$
Putting $y=1$ in (3), we get
$$f(f(z))=z f(1) \quad \forall z \in \mathbf{R}$$
Now using (2) and (4), we obtain
$$f(x f(x))=f\left(f\left(x^{2}\right)\right)=x^{2} f(1)$$
Put $y=z=x$ in (3) also given
$$f(x f(x))=x f(x)$$
Comparing (5) and (6), it follows that $x^{2} f(1)=x f(x)$. If $x \neq 0$, then $f(x)=c x$, for some constant $c$. Since $f(0)=0$, we have $f(x)=c x$ for $x=0$ as well. Substituting this in (1), we see that
$$c\left(x^{2}+c y z\right)=c x^{2}+c y z$$
or
$$c^{2} y z=c y z \quad \forall y, z \in \mathbf{R}$$
This implies that $c^{2}=c$. Hence $c=0$ or $1$. We obtain $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$. It is easy to verify that these two are solutions of the given equation.