Given that $f(x)+f(y) = f[x \times \sqrt{1-y^2}+ y\times \sqrt{1-x^2}]$ To prove $f(4x^3-3x) + 3f(x) = 0$ Clearly $\arcsin x$ is one such function but how do we prove it in general?
I differentiated both sides wrt x but that didn't give me something to get even close to what's required
 A: The functional equation makes sense only for $x,y\in[-1,1]$.
We rewrite the functional equation as 
$$\tag0 f(\sin t)+f(\sin u)=f(\sin(t+u))\qquad\text{for }-\frac\pi2\le t,u\le\frac\pi 2 $$
(the range restriction comes from the need to have $\sqrt{1-\sin^2}=\cos$). In other words, with $g=f\circ \sin\colon[-\pi,\pi]\to \Bbb R$, we have
$$\tag1 g(x)+g(y)=g(x+y)\qquad \text{for} -\frac\pi2\le x,y\le\frac\pi 2.$$
Preliminary remarks:
At least without the restriction on the domain, this is a well-known problem, but even with that constraint, we find:
$$ g(0)=0$$
(and so $f(0)=0$) by setting $x=y=0$; then
$$ \tag2g(-x)=-g(x)\qquad \text{for }|x|\le \frac\pi 2,$$
(and so $f(-x)=-f(x)$ for $x\in[-1,1] $). Next, by induction
$$ g(nx)=ng(x)\qquad\text{if }n\in\Bbb N, |(n-1)x|\le\frac\pi2.$$
Using this and $(2)$,
$$ \tag3g(ax)=ag(x)\qquad \text{if }a\in\Bbb Q, |x|\le\pi, |ax|\le\pi.$$
If we make the additional assumption that $g$ is continuous, this implies
$$ g(x)=x\cdot g(1)\qquad \text{for }|x|\le \pi$$
and consequently,
$$ f(x)=g(\arcsin x)=g(1)\cdot \arcsin x.$$
However, we cannot (and need not) assume that $g$ (or $f$) is continuous.
Nevertheless, $(3)$ and $(2)$ still give us
$$\begin{align}
-f(4\sin^3t-3\sin t)&=-f(-\sin 3t)\\
&=f(\sin 3t)\\
&=g(3t)\\
&=3g(t)\\
&=3f(\sin t)
\end{align} $$
provided $|t|\le \frac \pi 3$, hence
$$ \fbox{ $f(4x^3-3x)+3f(x)=0\qquad \text{if }|x|\le\frac{\sqrt 3}2.$ }$$
So we still ought to show the result for $\frac{\sqrt 3}2<|x|<1$, i.e., for $\frac \pi3<|t|\le \frac \pi2$.
Since $f$ is odd, we need only consider $\frac\pi3<t\le\frac \pi2$ and write $t=\frac\pi3+u$ with $0<u<\frac\pi 6$.
Now we get
$$\begin{align}
-f(4\sin^3t-3\sin t)&=-f(-\sin 3t)\\
&=-f(\sin 3u)\\
&=-g(3u)\\
&=-3g(u)\\
&=-3f(\sin u)
\end{align} $$
but unfortunately, we will in general not have $f(\sin u)=-f(\sin t)$.
A: Assuming functional equation must hold for every $x,y \in [-1, 1]$:
Set $x = y = 0$ to get $f(0) = 0$.
Set $y = 1$ to get $f(x) + f(1) = f(\sqrt{1-x^2})$, then apply the same with $-x$ instead of $x$ and compare: $f(x) + f(1) = f(-x) + f(1)$, hence $f(x)= f(-x)$.
Set $y = -x$ to get $f(x) + f(-x) = f(0) = 0$; combined with the above, $f(x) = 0$ for all $x$.
The desired result is now trivially true.
