The correct answer is $\boxed{\text{A. }\frac{1}{{{{\rm{x}}^2}}}}$.
A function is said to be bounded if it has an upper and lower bound. In other words, there are two numbers $M$ and $m$ such that $M \ge f(x) \ge m$ for all $x$ in the domain of $f$.
The function $\frac{1}{{{{\rm{x}}^2}}}$ is bounded because it approaches $0$ as $x$ approaches $\pm \infty$. In other words, there are two numbers $M$ and $m$ such that $M \ge \frac{1}{{{{\rm{x}}^2}}} \ge m$ for all $x$ in the domain of $\frac{1}{{{{\rm{x}}^2}}}$.
The function $e^x$ is not bounded because it approaches $\infty$ as $x$ approaches $\infty$. In other words, there is no number $M$ such that $M \ge e^x$ for all $x$ in the domain of $e^x$.
The function $x^2$ is not bounded because it approaches $\infty$ as $x$ approaches $\infty$. In other words, there is no number $M$ such that $M \ge x^2$ for all $x$ in the domain of $x^2$.
The function $e^{-x^2}$ is not bounded because it approaches $0$ as $x$ approaches $\pm \infty$. In other words, there is no number $M$ such that $M \ge e^{-x^2}$ for all $x$ in the domain of $e^{-x^2}$.