How to find the second derivative of the inverse function of $f(x)$ at $x=0$? If $f\displaystyle(x)=\int _{ \sin x }^{ \cos x }{ \frac { dt }{ e^{ t }\sqrt { 1-t^2 }  }  } $ where $x\in[0,\pi/2]$ then how to find the second derivative of the inverse function of $f(x)$ at $x=0$ ?

Denote the inverse function by $h(x)$ so that $h(f(x)) = x$. Using implicit differentiation, we have
$$ h'(f(x)) \cdot f'(x) = 1, \\
 h''(f(x)) \cdot (f'(x))^2 + h'(f(x)) \cdot f''(x) =  h''(f(x)) \cdot (f'(x))^2 + \frac{f''(x)}{f'(x)} = 0.$$
Note that $f(\frac{\pi}{4}) = 0$ so
$$ h''(0) = -\frac{f''(\frac{\pi}{4})}{(f'(\frac{\pi}{4}))^3}. $$
Calculate $f'(\frac{\pi}{4})$ and $f''(\frac{\pi}{4})$ using the fundamental theorem of calculus and you're done.

The function $f$ is of the form 
$$ f(x) = \int_{g(x)}^{r(x)} s(t) \, dt = \int_q^{r(x)} s(t) \, dt - \int_q^{g(x)} s(t) \, dt$$ where $q \in \mathbb{R}$ is some arbitrary point. Using the chain rule and the fundamental theorem of calculus, we see that
$$ f'(x) = s(r(x)) \cdot r'(x) - s(g(x)) \cdot g'(x). $$ 
Applying this to your specific $f$, we have
$$ f'(x) = \frac{-\sin(x)}{e^{\cos(x)}\sqrt{1 - \cos^2(x)}} - \frac{\cos(x)}{e^{\sin(x)} \sqrt{1 - \sin^2(x)}} = - \left( \frac{1}{e^{\cos(x)}} + \frac{1}{e^{\sin(x)}} \right) = -e^{-\cos(x)} - e^{-\sin(x)}. $$
