You try to make the RHS equal to zero.

cos(-pi/2) = 0

So take the cosines of both sides of your original equation.

Let A = arcsin(x/3)

And B = 2arctan[sqrt{(x+3)/(x-3)}]

So your original equation is in the form

A -B = -pi/2

Take the cosines of both sides,

cos(A-B) = 0

cosAcosB +sinAsinB = 0 ------------***

A = arcsin(x/3)

So,

sinA = x/3 ----------------(i)

cosA = sqrt(9 -x^2) /3 ------(ii)

B = 2arctan[sqrt{(x+3)/(x-3)}]

So,

sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])

sinB = sqrt(9-x^2) /18 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])

cosB = -x/18 ----------------------------(iv)

Hence,

cosAcosB +sinAsinB = 0 ------------***

[(sqrt(9 -x^2) /3)*(-x/18)] +[(x/3)*(sqrt(9 -x^2) /18)] =? 0

[-x*sqrt(9 -x^2) /54] +[x*sqrt(9 -x^2) /54 =? 0

0 =? 0

Yes.

Therefore, your original equation is proven.