Disclaimer: I am borrowing much from the Art of Problem Solving link alluded to in the comments.
Since you have guessed a solution, namely $x \mapsto \sqrt{1-x^2}$, so let us exploit it. Denote $s(x) := \sqrt{1-x^2}$, and note, as you did, that $s(s(x)) = x$. It is a good idea to express $f$ in an alternative way so that for $f(x) = s(x)$ things become very simple. This is vague, but a way to proceed is to write $f(x) = h(s(x))$ for some continuous $h:[0,1] \to \mathbb{R}$; since this is equivalent to saying that $h(y) = f(s(y))$, there is a 1 to 1 correspondence between possible $f$'s and $h$'s.
Let us see what the conditions say about $h$:
(i) $h(y) = f(s(y)) = \frac{1 + y^2}{2} f\left(\frac{1-y^2}{1+y^2}\right) = \frac{1 + y^2}{2}h\left(s\left(\frac{1-y^2}{1+y^2}\right)\right) = \frac{1 + y^2}{2} h\left( \frac{2y}{1+y^2} \right)$ (ii) $h(1) = 1$
(iii) $\lim_{y \to 0+} \frac{h(y)}{y}$ exists.
There are several ways to proceed at this stage. To keep things elegant, let us notice the striking resemblance between the expression $\frac{2y}{1+y^2}$ and the formula for $\tanh$ of doubled angle. In fact, if $y = \tanh \alpha$, then $\frac{2y}{1+y^2} = \tanh 2 \alpha$. Writing additionally $\frac{1+y^2}{2}$ as $\frac{y}{ \frac{2y}{1+y^2}} $ we conclude that (i) can be re-expressed in a nicer form: $ h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} h(\tanh 2\alpha)$ Iterating this as many times as we like, we conclude: $ h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} \frac{\tanh 2 \alpha}{\tanh 4 \alpha} \dots \frac{\tanh 2^{k-1} \alpha}{\tanh 2^{k} \alpha} h(\tanh 2^k\alpha) = \tanh \alpha \frac{h(\tanh 2^k\alpha)}{\tanh 2^k \alpha}$ Passing to the limit $k \to \infty$ (and remembering that $\tanh \beta \to 1 $ as $\beta \to \infty$ we conclude that: $ h(\tanh \alpha) = \tanh \alpha$ This means that $h(y) = y$. Translating this back to $f$ we conclude that: $ f(x) = h(s(x)) = s(x) = \sqrt{1-x^2}$