Inverse of ratio function $$ f(x) =\frac{4x^3}{x^2+1}$$

Question is 
$$ \frac d{dx} (f^{-1}(2))=? $$

Now i know how to invert exponential function, rational function and etc but i don't understand how can i invert this function

Try to exchange x and y 

So 
$$ x=\frac{4y^3}{y^2+1} $$

And then couldn't find out how to do
In General, because an inverse is the reflection of the function in the line $y=x$, the slope of the tangent to the inverse at the point $(x,f^{-1}(x))$ must be the reciprocal of the slope of the tangent to the function at $(f^{-1}(x) ,x)$
$$ 
\frac d{dx}f^{-1}(x)=\frac 1 {f'( f^{-1}(x))}
$$

In your case $f(1)=2$ so $f^{-1}(2)=1$ 

So
$$ 
\frac d{dx}f^{-1}(2)=\frac 1 {f'( 1)}
$$
The inverse function is:

$$\frac{1}{12} \left(\frac{x^2}{\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x}}+\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x}+x\right)$$

and its derivative is:

$$\frac{1}{12} \left(-\frac{\left(3 x^2+\frac{6 \sqrt{3} \left(2 x^3+2
   \left(x^2+108\right) x\right)}{\sqrt{x^2 \left(x^2+108\right)}}+216\right) x^2}{3
   \left(x^3+12 \sqrt{3} \sqrt{x^2 \left(x^2+108\right)}+216 x\right)^{4/3}}+\frac{2
   x}{\sqrt[3]{x^3+12 \sqrt{3} \sqrt{x^2 \left(x^2+108\right)}+216 x}}+\frac{3
   x^2+\frac{6 \sqrt{3} \left(2 x^3+2 \left(x^2+108\right) x\right)}{\sqrt{x^2
   \left(x^2+108\right)}}+216}{3 \left(x^3+12 \sqrt{3} \sqrt{x^2
   \left(x^2+108\right)}+216 x\right)^{2/3}}+1\right)$$

Here is the function (red), and its inverse (green) and the reflection line y = x (dashed blue):