A function is bijective if it is injective (one-to-one) and surjective (onto).
You can show $f$ is injective by showing that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
You can show $f$ is surjective by showing that for each $y \in \mathbb{R} - \{2\}$, there exists $x \in \mathbb{R} - \{-1\}$ such that $f(x) = y$.
If $f(x_1) = f(x_2)$, then
\begin{align*}
\frac{4x_1 + 3}{2x_1 + 2} & = \frac{4x_2 + 3}{2x_2 + 3}\\
(4x_1 + 3)(2x_2 + 2) & = (2x_1 + 2)(4x_2 + 3)\\
8x_1x_2 + 8x_1 + 6x_2 + 6 & = 8x_1x_2 + 6x_1 + 8x_2 + 6\\
8x_1 + 6x_2 & = 6x_1 + 8x_2\\
2x_1 & = 2x_2\\
x_1 & = x_2
\end{align*}
Thus, $f$ is injective.
Let $y \in \mathbb{R} - \{2\}$. We must show that there exists $x \in \mathbb{R} - \{-1\}$ such that $y = f(x)$. Suppose
$$y = \frac{4x + 3}{2x + 2}$$
Solving for $x$ yields
\begin{align*}
(2x + 2)y & = 4x + 3\\
2xy + 2y & = 4x + 3\\
2xy - 4x & = 3 - 2y\\
(2y - 4)x & = 3 - 2y\\
x & = \frac{3 - 2y}{2y - 4}
\end{align*}
which is defined for each $y \in \mathbb{R} - \{2\}$. Moreover, $x \in \mathbb{R} - \{-1\}$. To see this, suppose that
$$-1 = \frac{3 - 2y}{2y - 4}$$
Then
\begin{align*}
-2y + 4 & = 3 - 2y\\
4 & = 3
\end{align*}
which is a contradiction.
The inverse function is found by interchanging the roles of $x$ and $y$. Hence, the inverse is
$$y = \frac{3 - 2x}{2x - 4}$$
To verify the function
$$g(x) = \frac{3 - 2x}{2x - 4}$$
is the inverse, you must demonstrate that
\begin{align*}
(g \circ f)(x) & = x && \text{for each $x \in \mathbb{R} - \{-1\}$}\\
(f \circ g)(x) & = x && \text{for each $x \in \mathbb{R} - \{2\}$}
\end{align*}
\begin{align*}
(g \circ f)(x) & = g\left(\frac{4x + 3}{2x + 2}\right)\\
& = \frac{3 - 2\left(\dfrac{4x + 3}{2x + 2}\right)}{2\left(\dfrac{4x + 3}{2x + 2}\right) - 4}\\
& = \frac{3(2x + 2) - 2(4x + 3)}{2(4x + 3) - 4(2x + 2)}\\
& = \frac{6x + 6 - 8x - 6}{8x + 6 - 8x - 8}\\
& = \frac{-2x}{-2}\\
& = x\\
(f \circ g)(x) & = f\left(\frac{3 - 2x}{2x - 4}\right)\\
& = \frac{4\left(\dfrac{3 - 2x}{2x - 4}\right) + 3}{2\left(\dfrac{3 - 2x}{2x - 4}\right) + 2}\\
& = \frac{4(3 - 2x) + 3(2x - 4)}{2(3 - 2x) + 2(2x - 4)}\\
& = \frac{12 - 8x + 6x - 12}{6 - 4x + 4x - 8}\\
& = \frac{-2x}{-2}\\
& = x
\end{align*}
Hence, $g = f^{-1}$, as claimed.
That depends on subtleties of your definition of inverse function. If you expect a function $f\,:\, A \to B$ to have an inverse $f^{-1} \,:\, B \to A$, then $f$ needs to be surjective. If it is not, there are some $b \in B$ that aren't reached by $f$ at all, so how would you define $f^{-1}$ for such elements of $B$?
However, any injective $f\,:\, A \to B$ can be made surjective by simply restricting $B$, i.e. by setting $\widetilde{B} = \{f(a)\,:\, a\in A\}$ and redefining $f$ as $f \,:\, A \to \widetilde{B}$. That redefined $f$ then has an inverse $f^{-1} \,:\, \widetilde{B} \to A$.
Note, however, that while this makes the concatenation $f^{-1} \circ f$ the identity funtion on $A$, it does not make $f \circ f^{-1}$ the identity function on $B$, only on $\widetilde{B} \subsetneq B$.
Thus, if $f \,:\, A \to B $ is injective but not surjective, it has a left-inverse, because we can find an $f^{-1}$ as above with the property that $f^{-1} \circ f$ is the identity on $A$. But it does not have a right-inverse, because we won't find a $f^{-1}$ such that $f \circ f^{-1}$ is the identity on $B$ - the best we can achieve is an identity on $\widetilde{B}$.
Similarly, if $f \,:\, A \to B$ is surjective but not injective, it has a right-inverse, but no left-inverse.
Best Answer
It all depends on the co-domain of your function.
When you have a function $$f:A\to B$$ which is one-to-one but not onto $B$, you may restrict your co-domain to a subset of $B'\subset B$ which is the range of $f$.
For example $$f:N \to N $$ defined by $$f(n)=2n$$ is not onto but it is one-to-one.
If we define, $$f^* : N\to 2N$$ with the same definition $f^*(n)=2n$
We have an inverse function, $(f^*)^{-1} (n) = n/2.$