Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 28

You need to remember how Newton's method is used, to evaluate the roots of a transcendent equation.
x_(n+1) = x_n - (f(x_n))/(f'(x_n)) , where x_n is the approximate solution to the equation f(x) = 0.
You need to solve for x the equation e^(arctan x) = sqrt(x^3+1) , hence, you need to have the function in the form f(x) = 0 , such that:
e^(arctan x) - sqrt(x^3+1) = 0
Notice that for x = 0 yields:
e^(arctan 0) - sqrt(0+1) = e^0 - sqrt 1 = 1-1 = 0
You may consider the first approximation:
x_1 = 0 - (f(0))/(f'(0))
(f'(x)) = (e^(arctan x))/(1+x^2) - (3x^2)/(2sqrt(x^3+1))
(f'(0)) = (e^(arctan 0))/(1+0^2) - (0^2)/(2sqrt(0^3+1)) = 1 !+0
x_1 = 0 - 0/1 = 0
Hence, evaluating the root of the given transcendent equation, using Newton's method, yields x = 0.