Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right) $ 
If
$$f(x) +  f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$
prove that
$$f(4x^3 -3x) + 3f(x) =0\text.$$

I started by substituting $x = y$, in the expression and I get
$$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$
Then, I also substitute $y = \sqrt{(1-x^2)}$ and I get
$$f(x) +f\left(x\sqrt{1-x^2}\right) = 0\text.$$
How shall I proceed further and solve this problem?
 A: I'll assume the arguments of $f$ are always between $-1$ and $1$. Then
$$2f(x)=f(2x\sqrt{1-x^2})$$
and
$$3f(x)=f(x)+f(2x\sqrt{1-x^2})=f(y)$$
where
$$
\begin{split}
y &= x\sqrt{1-4x^2(1-x^2)}+\sqrt{1-x^2}2x\sqrt{1-x^2}\\
  &= x\sqrt{1-4x^2+4x^4}+2x\left(1-x^2\right)\\
  &= x\left(1-2x^2\right)+2x\left(1-x^2\right)=3x-4x^3.
\end{split}
$$
But $f(y)+f(-y)=f(0)=f(0)+f(0)$ and so $3f(x)=f(-y)$.
A: Easy way to prove that for every real number $ x $ with $ - 1 \le x \le 1 $, $ f ( x ) = 0 $ (using some steps already mentioned in the comments, but repeated here to be complete):
Let $ x = y = 0 $ in
$$ f ( x ) + f ( y ) = f \Big( x \sqrt { 1 - y ^ 2 } + y \sqrt { 1 - x ^ 2 } \Big) \tag { * } \label { * } $$
and you get $ f ( 0 ) = 0 $. Again, let $ x = y = 1 $ in \eqref{ * } and you get $ f ( 1 ) = 0 $. Thus, letting $ y = 1 $ in \eqref{ * } you'll have $ f ( x ) = f \big( \sqrt { 1 - x ^ 2 } \big) $. This also shows that $ f ( | x | ) = f ( x ) $. Now, substituting $ | x | $ for $ x $ and $ \sqrt { 1 - x ^ 2 } $ for $ y $ in \eqref{ * }, you get $ f ( x ) = 0 $.
