Let f be the real-valued function defined on the interval $(-1,1)$ such that $e^{-x}f(x)=2+ \int_0^x\sqrt{t^4+1}dt$ for all $x\in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then find the value of $(f^{-1})'(2)$.
This is my step
1-$f(x)=y$
2-$f^{-1}(y)=x$
3-$f^{-1}(x)=y$
4-$f^{-1}(x)'=\frac{dy}{dx}$
5- $y=f(x)=2e^{x}+e^{x} \int_0^x\sqrt{t^4+1}dt$
6- $y'=2e^{x}+e^{x} \int_0^x\sqrt(t^4+1)dt+e^x\sqrt{x^4+1}$
I am lost after this step, please help me as I have jolted down the step.
Clearly $f(0)=2$ and $$-e^{-x}f(x)+e^{-x}f'(x)=\sqrt{x^4+1}$$ at $x=0$ gives $-2+f'(0)=1$ then $$(f^{-1})'(2)=\dfrac{1}{f'(0)}=\dfrac13$$