Since the problem dos not provide the information where the Maclaurin's series is centered, you may consider that it is centered at x = 0, hence, evaluating the first three terms of Maclaurin's series yields:

`x/(sqrt(1 - x^2)) ~~ f(0) + (x - 0)*f'(0)/(1!) + (x - 0)^2*f''(0)/(2!) + ...`

You need to evaluate `f(0), f'(0), f''(0),` such that:

`f(0) = 0/(sqrt(1 - 0^2)) => f(0) = 0`

You need to evaluate f'(x) using quotient rule and chain rule, such that:

`f'(x) = (x'(sqrt(1 - x^2)) - x(sqrt(1 - x^2))')/(1 - x^2) `

`f'(x) = (sqrt(1 - x^2) + (2x^2)/(2sqrt(1 - x^2)))/(1 - x^2)`

`f'(x) = (sqrt(1 - x^2) + (x^2)/(sqrt(1 - x^2)))/(1 - x^2)`

`f'(x) = (1 - x^2 + x^2)/(sqrt(1 - x^2)^3)`

Reducing duplicate terms yields:

`f'(x) = 1 /(sqrt(1 - x^2)^3) => f'(0) = 1`

`f''(x) = -((3/2)(1 - x^2)^(3/2 - 1)(-2x))/((1 - x^2)^3)`

`f''(x) = ((3x)sqrt(1 - x^2))/((1 - x^2)^3) => f''(0) = 0`

`x/(sqrt(1 - x^2)) ~~ 0 + (x)*1 + 0 + ...`

Hence, evaluating the first three terms of Maclaurin's series yields `x/(sqrt(1 - x^2)) ~~ 0 + x + 0 +` ...