$\mathbf{R}^{*}$, such that for any non-zero real numbers $x, y$ satisfying $x^{2}+y \neq 0$, we have $f\left(x^{2}+y\right)=f^{2}(x)+\frac{f(x y)}{f(x)}$.
Parse: First, prove that $f(1)=1$. For this, let $f(1)=c$, in the condition
$$
f\left(x^{2}+y\right)=f(x)^{2}+\frac{f(x y)}{f(x)}
$$

set $x= \pm 1, y \notin\{-1,0\}$, we get $f(1+y)=f(1)^{2}+\frac{f(y)}{f(1)}$ (2), $f(1+y)=$ $f(-1)^{2}+\frac{f(-y)}{f(-1)}$ (3). In these two equations, set $y=1$, we have $f(2)=f(1)^{2}+1=f(-1)^{2}+1 \Rightarrow$ $f(-1)^{2}=f(+1)^{2} \Rightarrow f(-1)= \pm c$.
(1) If $f(-1)=c$, from (2) and (3) we get $\frac{f(y)}{f(1)}-\frac{f(-y)}{f(-1)} \Rightarrow f(y)=f(-y)$.
In (2), set $y=-2$, we get $c=f(-1)=f(1)^{2}+\frac{f(-2)}{f(1)}=c^{2}+\frac{f(2)}{c}=c^{2}+$ $\frac{c^{2}+1}{c} \Rightarrow c^{3}+1=0 \Rightarrow c=-1$. Thus, $f(2)=2, f(3)=f(1)^{2}+\frac{f(2)}{f(1)}=-1, f(4)=$ 2. But in (1), set $x=2, y=-2$, we get $2=f(2)=f(2)^{2}+\frac{f(-4)}{f(2)}=f(2)^{2}+\frac{f(4)}{f(2)}=$ $2^{2}+1=5$, a contradiction.
(2) If $f(-1)=-c$, similarly we get (using (2) and (3)): $f(y)=-f(-y)$.
Similarly, in (2), set $y=-2$, we get $-c=f(-1)=f(1)^{2}+\frac{f(-2)}{f(1)}=c^{2}-$ $\frac{f(2)}{f(1)}=c^{2}-\frac{c^{2}+1}{c}$, thus $c^{3}=1$, giving $c=1$. Therefore, $f(1)=1$.

Given $f(1)=1$, in (1), set $y=1$, we get $f\left(x^{2}+1\right)=f(x)^{2}+1$, and (2) becomes $f(1+$ $y)=1+f(y) \cdots$ (5). Therefore, $1+f(x)^{2}=f\left(1+x^{2}\right)=1+f\left(x^{2}\right)$, giving $f\left(x^{2}\right)=$ $f(x)^{2} \cdots$ (6), which indicates: for any $x>0$, $f(x)>0$. Furthermore, for any $x, y>0$, $f\left(x^{2}+y\right)=f(x)^{2}+\frac{f(x y)}{f(x)}>f(x)^{2}=f\left(x^{2}\right)$, which indicates: $f(x)$ is increasing on $(0,+\infty)$, and combining (4) we know $f(x)$ is increasing on $\mathbf{R}^{*}$.

Finally, we prove: for any $x \in \mathbf{Q} \backslash\{0\}$, $f(x)=x$, this way, combining $f(x)$ being increasing on $\mathbf{R}^{*}$, we know for any $x \in \mathbf{R}^{*}$, $f(x)=x$ (Cauchy method).

Using (5) and $f(1)-1$, combined with mathematical induction, we can prove: for any $x \in \mathbf{N}^{*}$, $f(x)=x$, and from (4) we know for any $x \in \mathbf{Z}^{-}$, $f(x)=x$, thus for any $x \in \mathbf{Z} \backslash\{0\}$, $f(x)=x$.
Now for any $x \in \mathbf{Q} \backslash\{0\}$, write $x=\frac{q}{p}, p \in \mathbf{N}^{*}, q \in \mathbf{Z},(p, q)=1, q \neq 0$, then $f\left(x^{2}+p\right)=f(x)^{2}+\frac{f(q)}{f(x)}$, using (5) and (6) we know $f\left(x^{2}+p\right)=f\left(x^{2}\right)+p=f(x)^{2}+p$, combining $f(q)-q$, we know $f(x)^{2}+p=f(x)^{2}+\frac{q}{f(x)} \Rightarrow f(x)=\frac{q}{p}=x$. Therefore, for any $x \in \mathbf{Q} \backslash\{0\}$, $f(x)=x$.
In summary, the function that meets the conditions is unique, which is $f(x)=x$.