Find $9$'th derivative of $\frac{x^3 e^{2x^2}}{(1-x^2)^2}$ How can I find $9$'th derivative at $0$ of $\displaystyle \frac{x^3 e^{2x^2}}{(1-x^2)^2}$. Is there any tricky way to do that?
This exercise comes from discrete mathematic's exam, so I think that tools like taylor  can't be used there.

Hint: Take $\ln$ of both sides of
$$y=\frac{x^3 e^{2x^2}}{(1-x^2)^2}$$
and use these facts that 
$$\dfrac{d^n}{dx^n}\ln(1+x)=\dfrac{(-1)^{n-1}(n-1)!}{(1+x)^{n+1}}$$
$$\dfrac{d^n}{dx^n}\ln(1-x)=\dfrac{(n-1)!}{(1-x)^{n+1}}$$
