<s>[INST] How to invert $f(x) = \frac{\sqrt{3}x}{\sqrt{1+3x^2}}$ Consider the function 
$f:(-1,1)\to(-1,1)$,
$f(x) = \frac{\sqrt{3}x}{\sqrt{1+3x^2}}$
To find $f^{-1}$ we have to show that f is injective and surjective.
Let $a,b\in(-1,1)$.
Then $f(a)=f(b)$ is equivalent to $\frac{\sqrt{3}a}{\sqrt{1+3a^2}}$=$\frac{\sqrt{3}b}{\sqrt{1+3b^2}}$.
It implies $(a-b)(a+b)=0$, which is problematic with $a=-b$.
To show surjectivity we have to show that for every $y$ there is an $x$ such that $f(x)=y,$ with which I'm also stuck!
 [/INST] Try first solving
$$
y=\frac{\sqrt{3}\,x}{\sqrt{1+3x^2}}
$$
First let's try it for $y>0$, so also $x>0$ (if it exists). Then we get
$$
y^2=\frac{3x^2}{1+3x^2}=1-\frac{1}{1+3x^2}
$$
that is
$$
\frac{1}{1+3x^2}=1-y^2
$$
This is possible only if $0<y<1$ and, in this case, we have
$$
1+3x^2=\frac{1}{1-y^2}
$$
so
$$
3x^2=\frac{1}{1-y^2}-1=\frac{y^2}{1-y^2}
$$
and so
$$
x=\frac{1}{\sqrt{3}}\frac{y}{\sqrt{1-y^2}}
$$
If $y\le0$ you find exactly the same formula under the condition $-1<y\le0$.
So $f$ is indeed invertible (over the range $(-1,1)$).
Note that this does show injectivity. However, if you're requested to go the hard way, here it is.
Suppose
$$
\frac{\sqrt{3}\,a}{\sqrt{1+3a^2}}
=
\frac{\sqrt{3}\,b}{\sqrt{1+3b^2}}
$$
Then either $a\ge 0$ and $b\ge0$ or $a<0$ and $b<0$.
Let's do the first case. Squaring we get
$$
\frac{3a^2}{1+3a^2}=\frac{3b^2}{1+3b^2}
$$
or
$$
1-\frac{1}{1+3a^2}=1-\frac{1}{1+3b^2}
$$
that implies
$$
1+3a^2=1+3b^2
$$
and therefore
$$
a^2=b^2
$$
so $a=b$.
Similarly for the case $a<0$ and $b<0$.
</s>