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.