Calculus · Hard problem

Let $f(x)=2\arctan x+\arctan\!\left(\dfrac{2x}{1-x^2}\right)$ for $0<x<1$. Using the identity $\arctan\!\left(\dfrac{2x}{1-x^2}\right)=2\arctan x$ (valid here), $f$ simplifies to $4\arctan x$. Find the value of $x$ at which $f'(x)=\dfrac{16}{5}$.