Denote by $f$ the function that maps a real number $x$ to $$f(x) = \arctan(\sqrt{x^2+1}-x)$$
Calculus shows that $f$ is a bijection from $\mathbb{R}$ to $\left(0,\dfrac{\pi}{2}\right)$
Now the question is to prove that the equation $f(x)=f^{-1}(x)$ has a unique solution in the interval $(0,1)$.
I'm aware that $f$'s expression can be simplified to $\dfrac{\pi}{4} - \dfrac{\arctan x }{2}$ but this is a question that comes afterwards in the problem.
I'm also aware that if $f$ is increasing on a given interval $I$ then the equation $f(x)=f^{-1}(x)$ (assuming all required conditions are fulfilled) is equivalent to $f(x) =x$ which is easier to solve (or deal with) in most cases. But $f$ here is decreasing.
$$f(x)=f^{-1}(x)\Leftrightarrow ff(x)-x\text{ has root in $(0,1)$}$$
Define $g(x)=ff(x)-x$.
$$g’(x)=\frac14\frac1{1+f^2(x)}\frac1{1+x^2}-1$$
For $x\in (0,1)$, we have $\frac\pi4\ge f(x)\ge\frac\pi8$ and $1>x>0$ which implies $g’(x)< 0$.
So $g(x)$ is decreasing.
$$g(0)=\frac\pi 4-\frac{\text{arctan}\frac\pi4}2>0$$ $$g(1)=\frac\pi4-1-\frac{\text{arctan}\frac\pi8}2<0$$
By intermediate value theorem $g$ has root in $(0,1)$, implying $f(x)=f^{-1}(x)$ has solution in $(0,1)$. Monotone of $g$ implies uniqueness of solution.