As you say, $f \,:\, A \to \mathbb{R}$ with $A \subset \mathbb{R}$ means that for every $x \in A$, the derivative $f'(x)$ exists. In other words, for every $x \in A$, the limit $$
f'(x) = \lim_{y \to x} \frac{f(y) - f(x)}{y-x}
$$
exists. Note that, due to the way limits works, whether or not $f'(x)$ exists depends only on the behaviour of $f$ in the immediate vicinity of $x$. More formally, if you have two functions $f$ and $g$, and some $\epsilon$ such that $f(y) = g(y)$ for all $y \in (x-\epsilon,x+\epsilon)$, then $f'(x) = g'(x)$.
Since you probably already know that the derivatives of $x \to x^2$ and $x \to -x^2$ exist on the whole real line, you therefore know that the derivative of $$
f(x) = \begin{cases} x^2 &\text{if $x\geq 0$} \\ -x^2 &\text{if $x < 0$} \end{cases}
$$
exists everywhere except at $x=0$. Because if $x \neq 0$, then $f(x) = x^2$ or $f(x) = -x^2$ on some small interval around $x$.
So all you have to do is to decide whether or not the derivate at $x=0$ exists or not. The first step is to check that $f$ is continuous at $x=0$. Since both $x^2$ and $-x^2$ are continuous $0$, it suffices to check that they take the same value at $x=0$ - and they do, $0^2 = -0^2$ after all. Note that being continuous at a point is a necessary (but not sufficient!) condition for being differentiable at that point. So had the continuity check failed, you could have immediately concluded that $f$ is not differentiable at $0$.
Since $f$ is continuous at $x=0$, and since the derivative of both $x^2$ and $-x^2$ also exists at $x=0$, it's sufficient for the derivative of $x^2$ and $-x^2$ to also take the same value at $x=0$ for $f$ to be differentiable there. And, indeed, they do - $\frac{d}{dx}(x^2) = 2x$ and $\frac{d}{dx}(-x^2) = -2x$, which at zero both take the value $0$. Thus, your $f$ is indeed differentiable everywhere.
The composition of two functions that are continuous everywhere is continuous everywhere. Since $|\cdot|$ and $x^2+2x−3$ are continuous everywhere, and your $f(x)$ is the composition of $|\cdot|$ and $x^2+2x−3$, you can conclude.
Best Answer
The following are theorems, which you should have seen proved, and should perhaps prove yourself:
Then you put together the parts. For example, $\frac 1x$ is continuous everywhere except perhaps at $x=0$, by point 6, because it is a quotient of a constant function (point 1) and the identity function (point 2). Then by point 4, $\cos \frac 1x$ is continuous everywhere except perhaps at $x=0$, because it is a composition of the cosine function, which is continuous for all $x\ne 0$, and the function $\frac 1x$. You can fill in the rest.