Let $f(x)=\frac{x^3}{x^2+1}$, and $g(x)$ is the inverse function of $f(x)$. Then $f(-1)=\frac{-1}{2}$ and $g(\frac{-1}{2})=-1$. Find $g'(\frac{-1}{2})$ and $g''(\frac{-1}{2})$.
I have found $f'(x)=\frac{x^4+3x^2}{(x^2+1)^2} $ and $f''(x)=-\frac{2x^3-6x}{(x^2+1)^3}$, but where do I go from there?
Hint: compute the derivatives of $g(f(x))=x$ and get $g'(f(x))f'(x)=1$ and $g''(f(x))(f'(x))^2 + g'(f(x))f''(x)=0$.