If a function's inverse exists, then the range of $f$ is the domain of $f^{-1}$, and vice-versa:
...because if a function's inverse exists, the function is then bijective: a one-to-one and onto function. Then the domain of the function is the "image" of it's inverse, and the range is the image of the domain.
NOTE: I'm not clear what you mean by the inverse function having the same form as the range. If you mean that to ask if domain of the inverse function is the range of the function, then yes, that's true, and that's what I'm addressing above. Otherwise, please clarify.
The procedure your professor used is a good tool for finding both the inverse function, if it exists, and for defining the inverse of the image of a function.
In your case, if you are asking if a function and its inverse (if it exists) have the same "form", you need to be clear about what you mean. The fact that both your $f(x)$ and $f^{-1}(x)$ are represented as the quotient of polynomials, and that meaning: "same form", then know, that will not always be the case. If we want to find the inverse of the image of $f(x) = x^2$, then $$y = x^2 \iff \pm\sqrt y = x \implies f^{-1}(x) = \pm\sqrt x$$
Here, we have $f(x) = x^2,\,$ and $\,f^{-1}(x) = \pm\sqrt x $
which hardly appear to be of the same "form" as far as functions go.
Consider the functions $f: \mathbb R^+ \to \mathbb R^+:
$$f(x) = e^x, \;\;f^{-1}(x) = \ln x$$
Would these functions be considered of the same "form"?
Feel free to give further clarification if your question is other than what's been addressed.
Best Answer
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.