Can you factor before finding derivative? Say the function is $y=\frac{x^2-1}{x-1}$
Can you factor functions before finding the derivative or does that not work?
 A: The functions $f(x) = \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1}$ and $g(x) = x+1$ are equal at every real number $x$ except $x = 1$, where $f(x)$ is undefined, but $g(x)$ is defined. 
Therefore, $f'(x) = g'(x) = 1$ at every real number $x$ except $x = 1$. 
If you used the quotient rule, you get 
$f'(x) = \dfrac{(x-1)(2x)-(x^2-1)(1)}{(x-1)^2} = \dfrac{x^2-2x+1}{(x-1)^2} = \dfrac{(x-1)^2}{(x-1)^2} = 1$ (provided $x \neq 1$). 
So, yes, you can factor and simplify the function before computing the derivative, but you must be careful to not enlarge the domain of the function or its derivative.
A: Yes, you can "factor" the numerator first, and cancel like terms, first before differentiating, but you'll need to exclude any values of $x$ such that $x$ is not defined in the original function.
For your initial example: $$\left(\frac{x^2 - 3}{x-3}\right)'$$ Yes, you can "factor" first before differentiating, but note $$(x^2-3) = (x - \sqrt 3)(x + \sqrt 3)\neq (x-3)(x+3).$$ So you're not going to gain any leverage by factoring the numerator first, since nothing will cancel. So taking the derivative of $$\frac{(x-\sqrt 3)(x+\sqrt 3)}{x - 3}$$ while you'll get the same exact result if you don't factor, will only complicate matters, since to differentiate the factored quotient will require both the product rule and quotient rule, or else the product rule for $3$ functions. 

In your newly posted example: $$\dfrac{x^2 - 1}{x - 1} = \frac{(x - 1)(x+1)}{x - 1}$$ Yes, indeed, you can (and I'd suggest you should, to simplify the task at hand) factor the numerator, cancel the common factor $(x - 1)$ which appears in both the numerator and denominator... 

but also need to make it very clear in your work that canceling the common factor is valid provided $x\neq 1$, and so neither the resulting function, nor its derivative, are defined at $x = 1$: e.g., $$y = \dfrac{x^2 -1}{x-1} = x+1 \iff x\neq 1$$ 

If you don't provide a comment to this effect, that information is lost after canceling, and thus incorrectly implying that , e.g., $f(1)$ and $f'(1)$ exist, when in reality they don't, given the original function. 
A: Short answer: Yes.
Factoring an expression preserves the function it represents.  That is, the functions
$$f(x) = (x+1)(x-1)$$
$$g(x) = x^2-1$$
are the same function.
Thus, taking the derivative of the factored form or the non-factored form is equivalent.
