How do you show that a function is differentiable if there isn't a specific $x_{0}$ given? Let's say we got a function given, simple example like $f(x)=x^{2}$. We are supposed to show this function is differentiable but there is no specific $x_{0}$ given where we check this.
What shall I do in this case?
Because I always used this formula to show it's differentiable:
$$\lim_{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$$
and check if for left and right side there is same solution, if so, it's differentiable.
But I cannot use this formula because there is no $x_{0}$.
What then? Can I just derivate the function and if it works, conclude it is differentiable? Oh and additionally show that there is no definition-gap, jump,..?
Please do tell me what to do in such a case, I wouldn't know what to do. I smell a task like that in my exam.

You can't just take the derivative because you're assuming what you're trying to prove. That being said, here's what we do:
As mentioned in my comment, we'll use the $x-c$ definition (you actually wrote it incorrectly in your question).
We say that $f$ is differentiable at $c$ if the limit
$$ \lim_{x \to c} \frac{f(x) - f(c)}{x-c} $$
exists. We call such a limit the derivative. Some books give this with $c$ relabeled as $x_0$ as you have done above.
So let $f(x) = x^2$. Then what? Well,
$$ \frac{x^2 - c^2}{x-c} = x+c $$
and now you can take the limit as $x \to c$. Done!
