Observe that
$$
\int \operatorname{d}^3\pmb r\; f(\pmb r)δ(\pmb r−\pmb r_0)=f(\pmb r_0)
$$
so that
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=\frac{1}{1+|\pmb r_0|^2}
$$
In spherical coordinates $(r,\theta,\phi)$, where $r \in [0,\infty)$, $\theta\in [0,\pi]$, and $\phi\in [0,2\pi)$,
$$ \delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{r^2}\frac{\delta(\theta-\theta_0)}{\sin^2\theta}\delta(\phi-\phi_0)\quad ~\textrm{Eq. 1 (cf. [1])}$$
and $\operatorname{d}^3\pmb r=r^2\sin^2\theta \operatorname{d}r\operatorname{d}\theta\operatorname{d}\phi$ so that
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{2\pi}\operatorname{d}\phi\,\delta(\phi-\phi_0)
\int_0^{\pi}\operatorname{d}\theta \delta(\theta-\theta_0)\int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0)
$$
then
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{\infty}\operatorname{d}r \frac{1}{1+r^2}\delta(r-r_0)=\frac{1}{1+r_0^2}
$$
Alternatively, observe that if the problem involves spherical coordinates, but with no dependence on either $\theta$ or $\phi$, one has
$$
\delta(\pmb r -\pmb r_0)=\frac{\delta(r-r_0)}{4\pi r^2}
$$
and
$$
\int \operatorname{d}^3\pmb r\frac{1}{1+\pmb r\cdot\pmb r}δ(\pmb r−\pmb r_0)=
\int_0^{2\pi}\operatorname{d}\phi
\int_0^{\pi}\operatorname{d}\theta \sin^2\theta
\int_0^{\infty}r^2\operatorname{d}r \frac{1}{1+r^2}\frac{\delta(r-r_0)}{4\pi r^2}=\frac{1}{1+r_0^2}
$$
Bibliography
[1] fen.bilkent.edu.tr/~ercelebi/mp03.pdf
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$$
\mbox{In spherical coordinates,}\quad
\delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad
\mbox{such that}
$$
\begin{align}
\color{#66f}{\large\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}}
&=\int_{0^{-}}^{\infty}\dd r\,r^{2}\int_{0}^{\pi}\dd\theta\,\sin\pars{\theta}
\int_{0}^{2\pi}\dd\phi\,{\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}
\\[3mm]&=\underbrace{\bracks{\int_{0^{-}}^{\infty}\delta\pars{r}\,\dd r}}
_{\ds{=\ 1}}\
\underbrace{\bracks{%
\int_{0}^{\pi}\delta\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta}}
_{\ds{=\ 1}}\
\underbrace{\bracks{\int_{0}^{2\pi}\delta\pars{\phi}\,\dd\phi}}_{\ds{=\ 1}}\
\\[3mm]&=\ \color{#66f}{\Large 1}
\end{align}
$$\mbox{Note that}\quad
\int_{{\mathbb R}^{3}}\delta\pars{\vec{r} - \vec{r}_{0}}\,\dd^{3}\vec{r}
=\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}
$$
Best Answer
The definition of the composition of the Dirac Delta, $\delta \circ f$, with a function $f$ assumes that $(1)$ $f$ is continuously differentiable and $(2)$ $f'$ is nowhere zero. Under such assumptions then if $f(x_i)=0$ for $i=1, \cdots N$ we define $\delta\circ f$ by
$$\delta \circ f=\sum_{i=1}^N \frac{\delta_{x_i}}{|f'(x_i)|}$$
where $\langle \delta_{x_i}, \phi \rangle =\phi(x_i)$ for any $\phi\in C_C^\infty$.
Then, we have
$$\langle \delta\circ f, \phi\rangle =\sum_{i=1}^N \frac{\phi(x_i)}{| f'(x_i)|}$$
Now, suppose $f(x)=\cos^2(x)-1$. Note that $f(0)=f(\pi)=0$. Note also $f'(x)=\sin(2x)$ so that $f'(0)=f'(\pi)=0$. Hence, $f(x)=\cos^2(x)-1$ fails to satisfy the assumptions of the definition.