Answer
$$ f^{(n)}(x)=(-1)^n (n+1)! \ x^{-(n+2)}.$$
Work Step by Step
Since $ f(x)=x^{-2}$, then we have
$$ f'(x)=(-2)x^{-3},\quad f''(x)=(-2)(-3)x^{-4}, \quad f'''(x)=(-2)(-3)(-4)x^{-5}, \cdots,$$
Hence, we get the general formula for $ f^{(n)}(x)$ as follows
$$ f^{(n)}(x)=(-1)^n (n+1)! \ x^{-(n+2)}.$$
