Solution:
Solution
Differentiating using Libnitz formula, we get
$$\sqrt{1-(f^{'}(x))^{2}}=f(x)$$
$$\sqrt{1-(f(x))^{2}}=f^{'}(x)=\cfrac{d(f(x))}{dx}$$
$$dx=\cfrac{d(f(x))}{\sqrt{1-(f(x))^{2}}}$$
Now, on integarting we get
$$x+C=\sin^{-1}(f(x))$$
Now for x to be real, we can say that
$$0\le f(x)<1 \, or\, f(x)<\cfrac{1}{2} and f(x)<\cfrac{1}{3}$$