If $f:X\to Y$ is a function from $X$ to $Y$, we usually call $X$ the domain, $Y$ the codomain, and $f(X)$ the range. Some authors call $Y$ the range, in which case $f(X)$ is called the image.
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
Since your function is bijective the domain of the inverse function is the codomain of the function and the codomain of inverse function is the domain of the function. Your formula for the inverse function is correct. $$ y=2x+3$$ implies $$x=(y-3)/2$$ Thus the inverse function is $$f^{-1} (x)=(x-3)/2.$$