1. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
f\left(x^{2}\right) f\left(y^{2}\right)+|x| f\left(-x y^{2}\right)=3|y| f\left(x^{2} y\right) \text {. }
$$
1. $f(x)=0$ and $f(x)=2|x|$.

Substituting $-x$ for $x$ in equation (1), we get
$$
f\left(x^{2}\right) f\left(y^{2}\right)+|x| f\left(x y^{2}\right)=3|y| f\left(x^{2} y\right) \text {. }
$$

Comparing with equation (1), we know that $f(x)$ is an even function.
Setting $x=y=0$ in equation (2), we get $f(0)=0$.
Swapping $x$ and $y$ in equation (2), we get
$$
f\left(x^{2}\right) f\left(y^{2}\right)+|y| f\left(x^{2} y\right)=3|x| f\left(x y^{2}\right) \text {. }
$$
Subtracting equation (3) from equation (2), we get
$$
|x| f\left(x y^{2}\right)=|y| f\left(x^{2} y\right) \text {. }
$$

Setting $y=1$ in equation (4), we get
$$
f\left(x^{2}\right)=|x| f(x) \text {. }
$$

Substituting equation (4) into equation (2), we get
$$
f\left(x^{2}\right) f\left(y^{2}\right)=2|x| f\left(x y^{2}\right) \text {. }
$$

Substituting equation (5) into equation (6), and for $x>0$, taking $y=\sqrt{x}$, we get
$$
f\left(x^{2}\right)=\frac{1}{2} f^{2}(x) \text {. }
$$

Since $f(x)$ is an even function and $f(0)=0$, equation (7) holds for any real number.
Combining with equation (5), we get
$$
\frac{1}{2} f^{2}(x)-|x| f(x)=0 \text {. }
$$

Solving, we find that for every real number $x$,
$$
f(x)=0 \text { or } f(x)=2|x| \text {. }
$$

If a non-zero real number $c$ satisfies $f(c)=0$, then by equation (5), we get $f\left(c^{2}\right)=0$.
Substituting into equation (6), we get that for any real number $y$, $f\left(c y^{2}\right)=0$.
Since $f(x)$ is an even function, $f(x)$ is always 0.
Therefore, only $f(x)=0$ and $f(x)=2|x|$ satisfy the conditions.