$f(x)=\frac{1}{x^{\frac{1}{3}}}$
Clearly it is visible that f(x) won't be defined at x=0($-1 \leq0 \leq 1$).
A function $f(x)$ is said to be continuous on a closed interval [a,b] iff
(i)f is continuous on the open interval (a,b)
(ii)$\lim_{x \to a^+}f(x)=f(a)$
(iii)$lim_{x \to b^-}f(x)=f(b)$
In words, f(x) is continous on [a,b] iff it is continous on (a,b) and it is continous at $a$ from the right and at $b$ from the left.
Since, the function f(x) here fails to satisfy point (1), it is not continous in closed interval [-1,1].For other intervals, you might need to check it.
A bounded function is one, which can be bounded if its set of values is bounded.
In short, it's graph can be bounded by using two lines y=k1 and y=k2 for some constants K1 and K2.
Here no need to check with graph, as $x \to 0$, $f(x) \to \infty$. Not Bounded ofcourse.
From question description it is given that
$A=\int_{-1}^{1}f(x)dx.$
Since, $f(x)$ is not defined at x=0, you need to integrate it piecewise.
$A=\int_{-1}^{0}f(x)dx+\int_{0}^{1}f(x)dx=3$
So, III is true.A is non-zero and finite.
So, option II and III true. (C)