Answer
$f$ is continuous everywhere.
Work Step by Step
$x=-1$ and $x=0$ are of interest to us, as they are the only values where $f(x)$ can have a discontinuity.
At $x=-1$:
Left-sided limit:$\qquad \displaystyle \lim_{x\rightarrow-1^{-}}f(x)=(-1)^{3}+2=1$
Right-sided limit:$\qquad \displaystyle \lim_{x\rightarrow-1^{+}}f(x)=(-1)^{2}=1$
The limit exists, $L=1$
Function value:$\quad f(-1)==(-1)^{3}+2=1$
So $f$ is continuous at $-1$.
At $x=0$:
Left-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 0^{-}}f(x)=0^{2}=0$
Right-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 0^{+}}f(x)=0$
The limit exists, $L=0..$
Function value:$\quad f(0)=0$
So $f$ is continuous at $0.$
Thus, $f$ is continuous everywhere.
