Sorry, I made a slight mistake in above answer. the correct answer is below.
Here, terms inside sqr root must be greater than or equal to 0.

1 – x
2 
0 or x
2 – 1

0 or (x+1)(x-1)

0
x

[-1, 1] …................(1)
1 –

(1 – x
2)

0 or

(1 – x
2)

1 or 1 – x
2 
1 or x
2 
0 , which is always true.
x

R ….......................(2)
1 –

1 –

(1 – x
2)

0 or 1 –

(1 – x
2)

1 or

(1 – x
2)

0
1 – x
2 
0 As i’ve written above x

[-1,1] ...................(3)
Take intersection of (1), (2) & (3) to get
x
[-1,1], which is the domain of the function