What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right)   \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right]   \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x   \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$), ($2$) and ($3$):
\begin{align}
3\arcsin x &= \arcsin \left[x\sqrt{1-2x\sqrt{1-x^2}^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-4x^2(1-x^2)}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-2(2x^2)+(2x^2)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{(2x^2-1)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x|2x^2-1|+ 2x(1-x^2)\right]
\end{align}
If $2x^2-1$ is positive, then $|2x^2-1|$ is $2x^2 -1$.
If $2x^2-1$ is negative, then $|2x^2-1|$ is $-2x^2+1$. 
Range of $x$ is $-1\leq x \leq 1  \implies  0\leq x^2 \leq 1  \implies  0\leq 2x^2 \leq 2$.
For $x\in\left(\frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$, then $2x^2-1$ is negative. 
For $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}\ , 1\right)$, then $2x^2-1$ is positive. 
Thus for  $x\in\left( \frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$
\begin{align} 
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin \left[-2x^3 +x+ 2x(1-x^2)\right]\\
&= \arcsin [3x - 4x^3]
\end{align}
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
\begin{align} 
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin [2x^3- x+2x-2x^3]\\
&= \arcsin[x] 
\end{align}
But clearly, $3\arcsin x = \arcsin[3x-4x^3]$.
So what is wrong whith this proof?
 A: For all positive $x$ (the case of negative $x$ is symmetric), $$\sin3x=3\sin x-4\sin^3x.$$
So with $x=\arcsin t$ we have
$$\sin(3\arcsin t)=3t-4t^3.$$
This allows us to write
$$3\arcsin t=\arcsin(3t-4t^3)\lor3\arcsin t=\pi-\arcsin(3t-4t^3).$$

As the range of the arc sine is $\left[0,\dfrac\pi2\right]$, the first identity holds up to $\arcsin t=\dfrac\pi6$, i.e. for $t\in\left[0,\dfrac12\right]$, then comes the second identity, for $t\in\left[\dfrac12,1\right]$.
$$3\arcsin t=\begin{cases}
t\le-\dfrac12&\to-\pi-\arcsin(3t-4t^3),
\\-\dfrac12\le t\le\dfrac12&\to\arcsin(3t-4t^3),
\\t\le\dfrac12&\to\pi-\arcsin(3t-4t^3).\end{cases}$$
A: A geometric point of view might be illuminating.
Suppose $0\leq x \leq \frac{1}{\sqrt 2}$. Consider the figure below, where we start from a right-angled triangle $\triangle ABC$ with sides $\overline{AB} = \sqrt{1-x^2}$ and $\overline{BC} = x$, and hypotenuse $\overline{AC}=1$. The choice of $x$ we have made guarantees that 
$$ \alpha = \angle BAC$$
is in the range $\left[0, \frac{\pi}{4}\right].$

By definition is
$$ \alpha = \arcsin x.$$
Extend first $CB$ to a segment $BD \cong BC$, then draw from $D$ the line perpendicular to $AD$ that intersects the extension of $AC$ in $E$. Finally draw from $C$ the perpendicular to $BC$ that meets $ED$ in $F$.
Define 
\begin{equation}\beta = \angle CAD = 2\alpha.\tag{1}\label{eq:1}\end{equation}
We have, by definition,
\begin{equation}\beta = \arcsin \left(\frac{\overline{ED}}{\overline{AE}}\right).\tag{2}\label{eq:2}\end{equation}
From $\triangle DCF \sim \triangle ABC$ find 
$$\overline{CF} = \frac{2x^2}{\sqrt{1-x^2}}$$
and
$$\overline{DF} = \frac{2x}{\sqrt{1-x^2}}.$$
By Pythagorean Theorem on $\triangle ADE$, and by $\triangle CEF \sim \triangle CED$
\begin{equation}
\begin{cases}
1+\overline{ED}^2 = (1+ \overline{EC})^2\\
\overline{EC} = \frac{x}{\sqrt{1-x^2}}\overline{ED}.
\end{cases}
\end{equation}
Solving the system yields
$$\overline{ED} = \frac{2x\sqrt{1-x^2}}{1-2x^2}$$
and
$$\overline{AC} = 1 + \overline{EC} = \frac{1}{1-2x^2}.$$
Using these results in \eqref{eq:2} and then cosidering the identity \eqref{eq:1} leads to
$$2\arcsin x = \arcsin \left(2x\sqrt{1-x^2}\right).$$

For $\frac{1}{\sqrt 2} \leq x \leq 1$, I would consider the triangle below, where again $\overline{BC} = x$, $\overline{AC} = 1$ and $D$ is the point symmetrical to $C$ with respect to line $AB$. Draw then from $C$ the perpendicular to $AD$ that meets its extension in $E$. Define then $\alpha$ as before and
$$\beta = \angle CAE = \pi - \angle CAD = \pi -2\alpha.$$
Use then the fact that $\sin \beta = \sin 2\alpha$. 

Finally, for negative $x$ just define $\overline{BC} = -x$ and proceed as above, taking advantage of the sine odd symmetry.
A: The problem is a conflict between these two lines:

$2\arcsin x= \arcsin(2x\sqrt{1-x^2}) \tag1$

and

Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$

The problem is that if $\frac1{\sqrt2} < \lvert x \rvert < 1,$
then $\frac\pi4 < \lvert\arcsin x\rvert < \frac\pi2,$ and therefore the left-hand side of equation $(1)$ obeys
$$ \frac\pi2 < \lvert2\arcsin x\rvert < \pi.$$
But the right-hand side of $(1)$ is just an application of the arc sine to a number, which must obey the conditions
$$ \lvert \arcsin (2x\sqrt{1-x^2}) \rvert \leq \frac\pi2.$$
Hence it is impossible for $(1)$ to be true when
$x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$
That means that everything you concluded by using $(1)$ -- which is to say, everything after the first few lines -- is inapplicable to the case
$x\in\left(-1, -\,\frac{1}{\sqrt2}\right) \cup \left(\frac{1}{\sqrt2}, 1\right).$
By using $(1),$ you restrict the validity of your argument to only the case
where $-\frac\pi4 \leq \arcsin x \leq \frac\pi4,$
that is, $x \in \left(-\,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right).$
The line

$ 3\arcsin x= \arcsin\left[x\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}+ 2x(1-x^2)\right] $

restricts the applicability of your argument even further, because again the right-hand side has magnitude at most $\frac\pi2,$ which means $\arcsin x$ can only have magnitude at most $\frac\pi6.$
So even the line

Thus for  $x\in\left(\frac{-1}{\sqrt2},\frac{+1}{\sqrt2}\right)$

is not right; everything you wrote after the first few lines is applicable only for
$-\frac12 \leq x \leq \frac12,$ so you really should have written,
"Thus for $x \in \left[-\frac12,\frac12\right]$".
For other $x$ you can follow Yves Daoust's answer.
