To prove that the function \(f(x)=\frac{1}{(x+1)^2}-2 x+\sin x\) has exactly one positive root, we'll use the Intermediate Value Theorem (IVT) and the properties of the function's first derivative. First, let's find the first derivative of the function.

\[

f^{\prime}(x)=\frac{d}{d x}\left[\frac{1}{(x+1)^2}-2 x+\sin x\right]

\]

Applying the differentiation rules, we get:

\[

f^{\prime}(x)=-\frac{2}{(x+1)^3}-2+\cos x

\]

Now, let's analyze the behavior of \(f(x)\) around \(x=0\). Evaluate \(f(0)\) and \(f^{\prime}(0)\) :

\[

\begin{gathered}

f(0)=\frac{1}{1^2}-2(0)+\sin 0=1 \\

f^{\prime}(0)=-\frac{2}{1^3}-2+\cos 0=-4+1=-3

\end{gathered}

\]

Since \(f(0)>0\) and \(f^{\prime}(0)<0\), we can deduce that the function is decreasing at \(x=0\).

Now, consider the limit of the function as \(x\) approaches infinity:

\[

\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}\left[\frac{1}{(x+1)^2}-2 x+\sin x\right]

\]

As \(x\) grows larger, the first and last terms of the function approach zero:

\[

\lim _{x \rightarrow \infty} \frac{1}{(x+1)^2}=0 \quad \text { and } \quad \lim _{x \rightarrow \infty} \sin x=-1 \leq \sin x \leq 1

\]

Thus, the limit becomes:

\[

\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty}(-2 x)=-\infty

\]

Since the function is continuous on the interval \((0, \infty)\), and we know that \(f(0)>0\) and \(\lim _{x \rightarrow \infty} f(x)=-\infty\), we can apply the Intermediate Value Theorem (IVT). The IVT states that for any value \(c\) between \(f(0)\) and \(\lim _{x \rightarrow \infty} f(x)\), there exists at least one point \(x_0\) in the interval \((0, \infty)\) such that \(f\left(x_0\right)=c\). In particular, since 0 is between \(f(0)\) and \(\lim _{x \rightarrow \infty} f(x)\), there must exist at least one positive root for the function \(f(x)\).

To show that there is exactly one positive root, we will now investigate the monotonic behavior of the function in the positive domain. If the function is either strictly increasing or strictly decreasing on an interval containing the root, it will ensure the uniqueness of the positive root. We know that \(f^{\prime}(x)=-\frac{2}{(x+1)^3}-2+\cos x\). Note that:

\[

-1 \leq \cos x \leq 1

\]

Therefore, we can find an upper bound for \(f^{\prime}(x)\) :

\[

f^{\prime}(x) \leq-\frac{2}{(x+1)^3}-2+1=-\frac{2}{(x+1)^3}-1

\]

Since the term \(\frac{2}{(x+1)^3}\) is always positive for \(x>0\), it follows that \(f^{\prime}(x)<0\) for all \(x>0\). This means that the function \(f(x)\) is strictly decreasing on the interval \((0, \infty)\).

Now, we have shown that:

1. There exists at least one positive root for \(f(x)\) by applying the Intermediate Value Theorem.

2. The function \(f(x)\) is strictly decreasing on the interval \((0, \infty)\).

Combining these two results, we can conclude that the function \(f(x)=\frac{1}{(x+1)^2}-2 x+\) \(\sin x\) has exactly one positive root, as the strictly decreasing behavior of the function ensures the uniqueness of the root.