Answer
$$
\frac{x}{x^{2}+1}=\sqrt {1-x}
$$
The root of the given equation, correct to eight decimal places, is 0.76682579.
Work Step by Step
$$
\frac{x}{x^{2}+1}=\sqrt {1-x}
$$
We first rewrite the equation in standard form:
$$
\frac{x}{x^{2}+1}-\sqrt {1-x}=0
$$
Therefore we let
$$
f(x)=\frac{x}{x^{2}+1}-\sqrt {1-x}=0.
$$
Then
$$
f^{'}(x)=\frac{1-x^{2}}{(x^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x}}
$$
so, Formula 2 (Newton’s method) becomes
$$
\begin{aligned} x_{n+1} &=x_{n}-\frac{f(x)}{f^{'}(x)}\\
\\ &=x_{n}-\frac{ \frac{x_{n}}{x_{n}^{2}+1}-\sqrt {1-x_{n}} }{\frac{1-x_{n}^{2}}{(x_{n}^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x_{n}}}}
\end{aligned}
$$
In order to guess a suitable value for $x_1$ we sketch the graphs of $y=\frac{x}{x^{2}+1} $ and $y = \sqrt {1-x} $ in the Figure .
It appears that they intersect at a point whose x-coordinate is somewhat less than 0.8, so let’s take $x_{1} = 0.8 $ as a convenient first approximation. Then Newton’s method gives
$$
\begin{aligned} x_{2} &=x_{1}-\frac{ \frac{x_{1}}{x_{1}^{2}+1}-\sqrt {1-x_{1}} }{\frac{1-x_{1}^{2}}{(x_{1}^{2}+1)^{2}}+\frac{1}{2\sqrt {1-x_{1}}}} \\
&=(0.8)-\frac{ \frac{(0.8)}{(0.8)^{2}+1}-\sqrt {1-(0.8)} }{\frac{1-(0.8)^{2}}{((0.8)^{2}+1)^{2}}+\frac{1}{2\sqrt {1-(0.8)}}} \approx 0.76757581
\end{aligned}
$$
repeating we get
$$
x_{3}\approx 0.76682610
x_{4} \approx 0.76682579 \approx x_{5} ,
$$
we conclude that the root of the given equation, correct to eight decimal places, is 0.76682579.