For $0<x<1,$ let $f(x)=\int_0^x \frac{dt}{\sqrt{1-t^2}}\ dt$

(1) Find $\frac{d}{dx} f(\sqrt{1-x^2})$

(2) Find $f\left(\frac{1}{\sqrt{2}}\right)$

(3) Prove that $f(x)+f(\sqrt{1-x^2})=\frac{\pi}{2}$
1. To find \(\frac{d}{dx} f(\sqrt{1-x^2})\):

   Given \( f(x) = \int_0^x \frac{dt}{\sqrt{1-t^2}} \), we recognize that this is the integral representation of the inverse sine function. Therefore, \( f(x) = \sin^{-1}(x) \).

   Now, we need to find the derivative of \( f(\sqrt{1-x^2}) \):
   \[
   \frac{d}{dx} f(\sqrt{1-x^2}) = \frac{d}{dx} \sin^{-1}(\sqrt{1-x^2})
   \]

   Using the chain rule:
   \[
   \frac{d}{dx} \sin^{-1}(\sqrt{1-x^2}) = \frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}} \cdot \frac{d}{dx} (\sqrt{1-x^2})
   \]

   Simplify the expression inside the square root:
   \[
   \sqrt{1-(\sqrt{1-x^2})^2} = \sqrt{1-(1-x^2)} = \sqrt{x^2} = |x| = x \quad \text{(since \(0 < x < 1\))}
   \]

   Now, find the derivative of \(\sqrt{1-x^2}\):
   \[
   \frac{d}{dx} (\sqrt{1-x^2}) = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1-x^2}}
   \]

   Putting it all together:
   \[
   \frac{d}{dx} \sin^{-1}(\sqrt{1-x^2}) = \frac{1}{x} \cdot \left( -\frac{x}{\sqrt{1-x^2}} \right) = -\frac{1}{\sqrt{1-x^2}}
   \]

2. To find \( f\left(\frac{1}{\sqrt{2}}\right) \):

   Since \( f(x) = \sin^{-1}(x) \):
   \[
   f\left(\frac{1}{\sqrt{2}}\right) = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)
   \]

   We know that:
   \[
   \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}
   \]

3. To prove that \( f(x) + f(\sqrt{1-x^2}) = \frac{\pi}{2} \):

   We have \( f(x) = \sin^{-1}(x) \) and \( f(\sqrt{1-x^2}) = \sin^{-1}(\sqrt{1-x^2}) \).

   We know that:
   \[
   \sin^{-1}(\sqrt{1-x^2}) = \cos^{-1}(x) \quad \text{for} \quad 0 < x < 1
   \]

   Therefore:
   \[
   f(x) + f(\sqrt{1-x^2}) = \sin^{-1}(x) + \cos^{-1}(x)
   \]

   Using the identity:
   \[
   \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}
   \]

   We conclude:
   \[
   f(x) + f(\sqrt{1-x^2}) = \frac{\pi}{2}
   \]

\(\blacksquare\)

The final answer is \(\boxed{\frac{\pi}{2}}\)