Prove that f is unique, and express f(x) in closed form. 
Let $f : [−1, 1] \to R$ be a continuous function such that
(i) $f(x) = \frac {2−x^2}{2}f(\frac {x^2}{2-x^2}) $ for every $x$ in [−1, 1],
(ii) $f(0) = 1$, and
(iii) $\lim_{x\to 1^{-}}\frac {f(x)}{\sqrt{1−x}}$ exists and is finite.
Prove that f is unique, and express f(x) in closed form.

Let $x=1$, we get $f(1)=0$. Let $\frac {x^2}{2-x^2}=1$, we have $x=\pm 1$, so let $x=-1$, we have $f(-1)=0$. It suffices to show that for any other function $g$ that satisfies the same conditions, $f(x)-g(x)=0$. We know that $f(x)-g(x)$ passes through the origin, thus we need to show that $f'(x)-g'(x)=0$. However, I don't know exactly what the third condition would imply. Anyone has a clue?

The third condition is actually a hint.
It implies that $\sqrt{1-x}$ is a factor of $f(x)$,
and it is not difficult to figure out that 
in order to fit the rest of conditions
the other factor should be $\sqrt{1+x}$. 
Indeed, the closed form expression is:
\begin{align}
f(x)&=\sqrt{1-x^2}.
\end{align}
Check on conditions:
\begin{align}
\checkmark
(\mathrm{i})& f(x)=\frac{2-x^2}{2}
\sqrt{1-\left(\frac{x^2}{2-x^2}\right)^2}
=\frac{1}{2}\sqrt{(2-x^2)^2-x^4}
=\sqrt{1-x^2}.
\\
\checkmark
(\mathrm{ii})& f(0)=1.
\\
\checkmark
(\mathrm{iii})& \lim_{x\to 1^{-}}\frac {f(x)}{\sqrt{1-x}}=\sqrt{2}.
\end{align}

