Showing a function is injective using that $f'(x)\ne0$
Yes, this suffices.
To see this suppose $f$ is differntiable on the domain of consideration but not injective. Then $f(a)=f(b)$ for some $a,b$ with $a\neq b$ and by the mean value theorem the derivative has a zero between $a$ and $b$.
Most of the time a function is not describable by a differentiable function so this method is used less often then by directly showing the claim. In algebra and general set theory for example this method is basically never applicable.
Yes, it is sufficient to show that $f'(x) \neq 0$ for all $x$.
One way to show the sufficiency is using Darboux's theorem. From this theorem, it follows that if we have $a,b$ such that $f'(a) > 0$ and $f'(b) < 0$, then there is a $c$ such that $f'(c) = 0$.
Thus, if we have $f'(x) \neq 0$ for all $x$, we may conclude that either $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$. Thus, $f'(x) > 0$ for all $x$, or $f'(x) < 0$ for all $x$, since we never have $f'(x) = 0$. The conclusion follows.
There's a little bit of subtlety. The mean value property will work if you can conclude that $f'$ is either always positive or always negative. But how do you conclude this from $f' \neq 0$?
If the derivative is assumed continuous, that will suffice, because if you know that $f'$ is positive at some point, and never zero, then it must remain positive, by the intermediate value theorem applied to $f'$.
But even if you don't assume $f'$ is continuous, you can still conclude $f'$ remains strictly positive or strictly negative, because a derivative always has the intermediate value property, namely:
If $f'(x) = a$ and $f'(y) = b$, and $a<c<b$, then there is some point between $x$ and $y$ where $f'(x)=c$.
That is true even if $f'$ is not continuous. To see how one can hold but not the other, consider the function $$x \mapsto \begin{cases} \sin (1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x =0, \end{cases}$$ which has the intermediate value property but is not continuous at zero*. This allows you to conclude that if $f'$ is never zero, it must remain strictly positive or strictly negative.
*And in fact it is the derivative of a function, namely $x^2 \cos(1/x) - 2 \int_0^x \xi \cos(1/\xi) \, d\xi$.