<s>[INST] Stuck with a tricky existence proof Show that there exists a continuous function $f: [-1, 1] \rightarrow \mathbb{R}$ such
$f(0) = 1$ and
$f(x) = \frac{2-x^2}{2} \cdot f(\frac{x^2}{2-x^2})$   
$\forall x \in [-1, 1]$
I tried putting in $x = 1$ and $x = -1$ in the second condition to find that $f(1) = f(-1) = 0$. 
I also took the derivative of the second equation to find that:
$f'(x) = x (f'(\frac{x^2}{2-x^2})\frac{2}{2-x^2}-f(\frac{x^2}{2-x^2}))$
This gives me
$f'(0) = f'(1) = f'(-1) = 0$ but now I'm stuck. Anybody see a way?
 [/INST] I recall solving this problem before... was it on a Putnam? Sadly, I don't remember what leap of intuition led me to the answer.
One may take $f(x) = \sqrt{1-x^2}$. It is clear that $f$ is continuous on $[-1,1]$ and that $f(0) = 1$, and one sees that
$$\frac{2-x^2}{2} \cdot \sqrt{1-\frac{x^4}{(2-x^2)^2}} = \frac{2-x^2}{2} \cdot \sqrt{\frac{x^4-4x^2+4-x^4}{x^4-4x^2+4}} \\
= \frac{2-x^2}{2} \cdot \sqrt{\frac{4-4x^2}{(2-x^2)^2}} = \sqrt{1-x^2},$$
so that $f(x) = \frac{2-x^2}{2} f(\frac{x^2}{2-x^2})$.
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