First, note that $g'(x)=1$ for $x>1$.
On the interior of each of the intervals $(-\infty,-1)$, $[-1,1]$, and $(1,\infty)$ $g$ is differentiable since each component function is. The only question is what happens at the endpoints of these intervals.
At $x=-1$, the value of both component functions is $1$ and the derivative of both component functions is $2$. This means we get
$$
\lim_{h\to0^-}\frac{g(-1+h)-g(-1)}{h}=-2\tag{1}
$$
Because $x^2=-1-2x$ at $x=-1$, we can use $g(x)=-1-2x$ for the computation of $(1)$.
Furthermore, we get
$$
\lim_{h\to0^+}\frac{g(-1+h)-g(-1)}{h}=-2\tag{2}
$$
using $g(x)=x^2$ for the computation of $(2)$.
Since the derivatives computed in $(1)$ and $(2)$ are the same, we get that $g(x)$ is differentiable at $x=-1$.
At $x=1$, the value of both component functions is $1$, however, the derivative of $x^2$ is $2$ and the derivative of $x$ is $1$. This means that
$$
\lim_{h\to0^-}\frac{g(1+h)-g(1)}{h}=2\tag{3}
$$
using $g(x)=x^2$ for the computation of $(3)$.
However, we get
$$
\lim_{h\to0^+}\frac{g(1+h)-g(1)}{h}=1\tag{4}
$$
Because $x=x^2$ at $x=1$, we can use $g(x)=x$ for the computation of $(4)$.
Since the derivatives computed in $(3)$ and $(4)$ are different, $\lim\limits_{h\to0}\frac{g(1+h)-g(1)}{h}$ does not exist, and therefore $g(x)$ is not differentiable at $x=1$.
If $f$ was differentiable at $a$, then $f'(a)=P'(a)=Q'(a)$. That's because, assuming that the derivative exists, $$f'(a)=\lim_{h\to0^-} \frac{f(a+h)-f(a)}{h}=\lim_{h\to0^-} \frac{P(a+h)-P(a)}{h}=P'(a)=\lim_{x\to a^-}P'(x)=\lim_{x\to a^-}f'(x)$$
and, since $P(a)=Q(a)=f(a)$,
$$f'(a)=\lim_{h\to0^-} \frac{f(a+h)-f(a)}{h}=\lim_{h\to0^+} \frac{Q(a+h)-Q(a)}{h}=Q'(a)=\lim_{x\to a^+}Q'(x)=\lim_{x\to a^+}f'(x)$$
and this implies that $\lim_{x\to a^-}f'(x)=\lim_{x\to a^+}f'(x)$.
Therefore if $\lim_{x\rightarrow{a}}f'(x)$ does not exist, then $f$ cannot be differentiable at $a$.
Observe that this is actually true whenever $P$ is continiously differentiable on $(a-\epsilon, a]$ and $Q$ is continiously differentiable on $[a,a+\epsilon)$, even without $P$ and $Q$ being polynomials
Best Answer
Taking $\lim\limits_{x\to 1}f'(x)$ is not well defined, until we are convinced that $f'(x)$ exists. If $f'(1)$ does exist, we will have that the left and right hand limits defining the derivative are equal. If we denote $f'_-(x)$ and $f'_+(x)$ as the left and right derivatives respectively, we get that \begin{align*} f'_-(x) &= 2\\ f'_+(x) &=\frac{1}{2} \end{align*} The left and right derivatives exist, but they are not equal. Thus, $f$ is not differentiable.