If $f: \mathbb R\setminus\{0\} \to \mathbb R$ be defined by $f(x)$ = $3 - \frac{2}{x}$. I'm trying to determine if the function is injective or surjective. I'm not sure if what I've done is completely correct.
Here's my attempt: Let $a, b \in \mathbb R\setminus\{0\}$, suppose $f(a) = f(b)$. Then
$$3 - \frac{2}{a}= 3 - \frac{2}{b} \Rightarrow \frac{2}{a} = \frac{-2}{a}\Rightarrow a= -b.$$
Therefore, $f$ is not injective.
To check surjectivity, let $y = 3 - \frac{2}{x}$, then
$$xy = 3- 2x \Rightarrow xy + 2x = 3 \Rightarrow x(y + 2) = 3 \Rightarrow x = \frac{3}{y} + 3.$$
Thus $f$ is not onto. ${}$
For injectivity:
From $$3-\frac{2}a=3-\frac{2}b$$
As you cancel $3$ from boths sides, we have
$$\color{blue}-\frac{2}a=-\frac{2}b$$
Try to examine if it is injective again.
Now for surjectivity, your goal is given $y\in R$, you want to check if you can find a non-zero preimage. If you can show a real value that it can't attain, then you have shown that it is not surjective.
Also, from your working, from $$y=3-\frac{2}x$$
If you multiply everything by $x$, you get
$$xy=\color{blue}{3x-2}$$
Can you see what value $y$ can't take? (it is a small positive integer number)