We have, using the identity $ \cos 2\theta=1-2\sin^2\theta$, that
$$\tag{1}\cos\bigl( 2\,{\rm arccot}\,{\textstyle{\sqrt x\over\sqrt y}}\,\bigr)= 1-2\,\sin^2\bigl( {\rm arccot}{\textstyle{\sqrt x\over\sqrt y}}\,\bigr).$$
To simplify
$\sin \bigl( {\rm arccot}{\textstyle{\sqrt x\over\sqrt y}}\bigr)$, let
$\theta= {\rm arccot}{\sqrt x\over\sqrt y}$ and draw a right triangle with one non-right angle $\theta= {\rm arccot}{\sqrt x\over\sqrt y}$. Then the cotangent of $\theta$ is $\sqrt x/\sqrt y$. Thus, we may take the length of the side of the triangle adjacent to $\theta$ to be $\sqrt x$ and the length of the side of the triangle opposite to $\theta$ to be $\sqrt y$. The Pythagorean Theorem then gives the length of the hypotenuse of the triangle to be ${ \sqrt{x+y}}$. Now, reading from the triangle, we have
$$\tag{2}
\sin \bigl( {\rm arccot}{\textstyle{\sqrt x\over\sqrt y}}\bigr)= {\sqrt y\over \sqrt{x+y}}
$$
Substituting $(2)$ into $(1)$ gives:
$$\eqalign{
\cos\bigl( 2\,{\rm arccot}\,{\textstyle{\sqrt x\over\sqrt y}}\,\bigr)= 1-2\,\sin^2\bigl( {\rm arccot}{\textstyle{\sqrt x\over\sqrt y}}\,\bigr)
&= 1-2\Bigl[{ {\sqrt y\over \sqrt{x+y}}}\Bigr]^2\cr
&={x+y\over x+y}-{2y\over x+y}\cr
&={x-y\over x+y}.
}
$$
This establishes the identity since
$$
\cos\bigl( {\rm arccos}\,{\textstyle{x-y\over x+ y}}\,\bigr) ={x-y\over x+y}.
$$
(The cosine function is
one-to-one on $[0,\pi]$, ${\rm arccos}\,{\textstyle{x-y\over x+ y}}\in[0,\pi]$, and
$2\, {\rm arccot}{\textstyle{\sqrt x\over\sqrt y}}\in(0,\pi ]$.)