Problem involving Fundamental Theorem of Calculus
The domain of the function $f(x) = \displaystyle \int_x^{x^2} \dfrac{t}{\log t} \, dt$ is $(0,1) \cup (1,\infty)$, but it would certainly be consistent to define $f(1) = 0$ since that integral is defined on a degenerate interval. Take the domain to be $(0,1) \cup (1,\infty)$.
Now, let $F$ be an antiderivative of $\dfrac{t}{\log t}$ defined on either interval $(0,1)$ or $(1,\infty)$. The fundamental theorem of calculus gives you $$ f(x) = F(x^2) - F(x)$$ so that $$f'(x) = F'(x^2) \cdot 2x - F'(x) = \frac{x^2}{\log (x^2)} \cdot 2x - \frac{x}{\log x}.$$
It is probably implicitly assumed that $x \in (0,\infty)\backslash \{1\}$. (Note that this guarantees that both $x$ and $x^2$ are either in $(0,1)$ or in $(1,\infty)$.)
You can split the integral into two parts, i.e. you can write $$ \int_x^{x^2} \frac{t}{\log t} \, dt = \int_x^{a} \frac{t}{\log t} \, dt + \int_a^{x^2} \frac{t}{\log t} \, dt = \int_a^{x^2} \frac{t}{\log t} - \int_a^{x} \frac{t}{\log t} , $$ where $a$ is a constant between $x$ and $x^2$, and then use the FTC on each term separately.
Notice that the integrand is only defined for $\mathbb R^+\backslash\{1\}$. Therefore the region $(x, x^2)$ never crosses $1$.
To your second question, we can always split the integral anywhere we like. For example, when $x \in (3,9)$,
$$ \int^{x^2}_x \frac{t}{\log t} \mathrm dt=\int^{x^2}_9 \frac{t}{\log t} \mathrm dt+\int^{9}_x \frac{t}{\log t} \mathrm dt $$
... and you can proceed by differentiating them seperately. When $x$ is not in this region, you can always choose some other splitting point, and sew the results together.