such questions should be dealt with a proper approach

merely looking at the question and guessing can be misleading at times

though in this case answer came correct by chance

but if it has come correct , its well and good

 

f(x)+f(y)=f [x.rt(1-y²)+y.rt(1-x²) ]

put x = y

f [ 2x.rt(1-x²) ] = 2f(x)

add f(x) on both sides

LHS f(x) + f [ 2x.rt(1-x²) ] = f ( 3x-4x³)

RHS = 3f(x)

so k = 3

or

put y = 2xroot (1-x²)

this directly gives the above equation

 

question is done.

but it can be proved (for more) that it is sin^-1r function

put x = y = 0

2f(0) = f(0)

so f(0) = 0....(1)

put y = -x

f(x) + f(-x)= f(0) = 0 from (1)

so f(-x) = -f(x)

so its an odd function