$\begin{array}{l}\text { 1. Let } f^{(1)}(x)=f(x), \\ f^{(n)}(x)=\left(f^{(n-1)}(x)\right)^{\prime}(n \geqslant 2) . \\ \text { If } f(x)=\frac{x^{2}}{1-x^{2}} \text {, then } f^{(n)}(0)=\square \text {. }\end{array}$
- 1. $\frac{1+(-1)^{n}}{2} \cdot n!$.

From the problem, we have
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{1-x^{2}}=-1+\frac{1}{1-x^{2}} \\
=-1+\frac{1}{(1-x)(1+x)} \\
=-1+\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right) . \\
\text { Then } f^{(n)}(x)=\frac{1}{2}\left(\frac{(-1)^{n} \cdot n!}{(1+x)^{n+1}}+\frac{n!}{(1-x)^{n+1}}\right) . \\
\text { Therefore } f^{(n)}(0)=\frac{1+(-1)^{n}}{2} \cdot n!.
\end{array}
$$