find properties(domain , even , surjective , injective) and asymptotes of a given function 
Given the function $f:\Bbb R \to \Bbb R$:
$
f(x)=
\begin{cases}
 \frac{\pi^2 -x^2}{x}&\text{if}\, -\infty<x<-\pi\\
 \pi^2-x^2&\text{if}\, -\pi\leq x<0\\
 1&\text{if}\, x=0\\
 \pi^2-x^2&\text{if}\, 0< x\leq \pi\\
 \frac{1}{x}\cdot \sin x &\text{if}\, \pi< x< \infty\\     
\end{cases}
$

*

*Find the domain

*Check if the function is injective , surjective and if the function is even

*Find the asymptotes in case they exist.




*

*The function is set for all real numbers since it is even defined for $x=0$ so the domain is $D=\{x \in \Bbb R\}$

*I did not understand how to approach this because there are different cases, do I approach each one alone?

If so,
Evenness: $\frac{\pi^2 -(-x)^2}{-x}\not=\frac{\pi^2 -x^2}{x}$ it is not even , but it is even in $-\pi \leq x <0$ and $0<x 
\leq \pi$ and also not even for $\frac{1}{x}\cdot \sin x$ as $\sin x$ is an odd function
Injective: for $\pi^2-x^2$ it is not injective because it is even  and $\frac{1}{x}\cdot sinx$ is not injective because $sinx$ is not injective , I am not sure but $\frac{\pi^2 -x^2}{x}$ is injective
surjective: I think it is because for every $f(x)$ we have $f(x)=y$
Asymptotes I checked every limit in the function
$\lim_\limits{x \to -\infty} \frac{\pi^2 -x^2}{x}=\infty$
$\lim_\limits{x \to -\pi^-} \frac{\pi^2 -x^2}{x}=0$
$\lim_\limits{x \to -\pi^+} \pi^2-x^2=0$
$\lim_\limits{x \to 0^-} \pi^2-x^2=\pi^2$
$\lim_\limits{x \to 0^+} \pi^2-x^2=\pi^2$
$\lim_\limits{x \to \pi^-} \pi^2-x^2=0$
$\lim_\limits{x \to \pi^+} \frac{\sin x}{x}=0$
$\lim_\limits{x \to \infty} \frac{\sin x}{x}=0$
From the last limit we can see that there is vertical asymptote  as $y=0$ (Not sure if it is considered one if $y=0$)
For oblique asymptote:
$a=\lim_\limits{x \to \infty} \frac{f(x)}{x}=\frac{\sin x}{x^2}=0$
$b=\lim_\limits{x \to \infty} f(x)-ax0$
No oblique asymptotes as well
Hopefully the translation are understandable.
I would like to know if my approach is correct, if yes then is there a different way?
If my approach is wrong then how do I approach these type of questions?
Thank you!

You are correct that the function is not even.  However, to show that the function is not even, you should demonstrate that there exist some $x \in \mathbb{R}$ such that $f(x) \neq f(-x)$.  For instance, $$f(2\pi) = \frac{\sin(2\pi)}{2\pi} = 0 \neq \frac{3\pi}{2} = \frac{-3\pi^2}{-2\pi} = \frac{\pi^2 - 4\pi^2}{-2\pi} = \frac{\pi^2 - (-2\pi)^2}{-2\pi} = f(-2\pi)$$
Your argument that
$$\frac{\pi^2 - x^2}{x} \neq \frac{\pi^2 - (-x)^2}{-x}$$
is not valid since the definition $f(x) = \frac{\pi^2 - x^2}{x}$ applies in the interval $(-\infty, -\pi)$ but not in the interval $(\pi, \infty)$.  Similarly, the definition $f(x) = \frac{\sin x}{x}$ applies in the interval $(\pi, \infty)$ but not in the interval $(-\infty, -\pi)$.
To show the function is not injective, you should produce two distinct elements in the domain with the same image.  For instance,
$$f\left(-\frac{\pi}{2}\right) = \pi^2 - \left(-\frac{\pi}{2}\right)^2 = \pi^2 - \frac{\pi^2}{4} = \frac{3\pi^2}{4} = \pi^2 - \frac{\pi^2}{4} = \pi^2 - \left(\frac{\pi}{2}\right)^2 = f\left(\frac{\pi}{2}\right)$$
or
$$f(2\pi) = \frac{\sin(2\pi)}{2\pi} = 0 = \frac{\sin(3\pi)}{3\pi} = f(3\pi)$$
To prove that the function is surjective, you must show that given $y \in \mathbb{R}$, there exists $x \in \mathbb{R}$ such that $f(x) = y$.  Notice that the function does not approach $-\infty$.  That suggests that it is bounded below.  In the interval $(-\infty, \pi)$,
$$f(x) = \frac{\pi^2 - x^2}{x} > 0$$
since both the numerator and denominator are negative.
In the interval $[-\pi, 0)$, $f(x) = \pi^2 - x^2 = (\pi + x)(\pi - x) \geq 0$, with equality holding if and only if $x = -\pi$ since both factors are positive if $-\pi < x < 0$ and $\pi + x = 0$ if $x = \pi$.
Clearly, $f(0) = 1 > 0$.
In the interval $(0, \pi]$, $f(x) = \pi^2 - x^2 = (\pi + x)(\pi - x) \geq 0$ since both factors are positive if $0 < x < \pi$ and $\pi - x = 0$ if $x = \pi$.
In the interval $(\pi, \infty)$,
$$f(x) = \frac{\sin x}{x} \geq \frac{\sin x}{\pi} \geq -\frac{1}{\pi}$$
since $x \geq \pi$ and $\sin x \geq -1$.
Thus, the function is bounded below by $-1/\pi$.  Hence, there is no value of $x$ in the domain such that $f(x) = -1$, so the function is not surjective.
A function has a vertical asymptote $x = a$ if $a \in \mathbb{R}$ and at least one of the following four statements is true:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty\\
\lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty\\
\lim_{x \to a^-} f(x) & = -\infty
\end{align*}
You have correctly shown that the function does not approach infinity or negative infinity at any finite value of $x$, so the function does not have a vertical asymptote.
A function has a horizontal asymptote $y = b$ if at least one of the following statements is true:
\begin{align*}
\lim_{x \to \infty} f(x) & = b\\
\lim_{x \to -\infty} f(x) & = b
\end{align*}
You have correctly shown that
$$\lim_{x \to \infty} f(x) = 0$$
Thus, the line $y = 0$ is a horizontal asymptote of the graph of $f$.
A function has an oblique asymptote $y = mx + b$, with $m \neq 0$, if at least one of the following statements is true:
\begin{align*}
\lim_{x \to \infty} [f(x) - (mx + b)] & = 0\\
\lim_{x \to -\infty} [f(x) - (mx + b)] & = 0
\end{align*}
Observe that in the interval $(-\infty, -\pi)$,
$$f(x) = \frac{\pi^2 - x^2}{x} = \frac{\pi^2}{x} - x$$
As $x \to -\infty$, the term $\pi^2/x$ approaches $0$, while the term $-x$ approaches infinity.  That suggests that the line $y = -x$ may be an oblique asymptote.  Indeed, it is since
\begin{align*}
\lim_{x \to -\infty} [f(x) - x] & = \lim_{x \to -\infty} \left[\frac{\pi^2 - x^2}{x} - (-x)\right]\\
& = \lim_{x \to -\infty} \left[\frac{\pi^2 - x^2}{x} + x\right]\\
& = \lim_{x \to -\infty} \frac{\pi^2 - x^2 + x^2}{x}\\
& = \lim_{x \to -\infty} \frac{\pi^2}{x}\\
& = 0
\end{align*}
