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$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) = \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no, even if the same norm is used on the domain and the co-domain of the linear operator corresponding to the matrix $A(x)$. (If different norms are allowed to be used for the domain and the co-domain, then the no answer is quite obvious.)

For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.

$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) := \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no, even if the same norm is used on the domain and the co-domain of the linear operator corresponding to the matrix $A(x)$. (If different norms are allowed to be used for the domain and the co-domain, then the no answer is quite obvious.)

For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.

$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) = \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no. For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.

$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) = \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no, even if the same norm is used on the domain and the co-domain of the linear operator corresponding to the matrix $A(x)$. (If different norms are allowed to be used for the domain and the co-domain, then the no answer is quite obvious.)

For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.

$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) = \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no. For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) = \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no. For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.