Answer
$f(x)$ is not continuous at $1$.
Work Step by Step
When $f(1)=\lim\limits_{x\to 1}f(x)$, then $f(x)$ will be continuous at $x=1$.
$\lim\limits_{x\to 1^+}f(x)=\lim\limits_{x\to 1^+}\dfrac{x^3-1}{x^2-1} \\=\lim\limits_{x\to 1^+}\dfrac{(x-1)(x^2+x+1)}{(x-1)(x+1)}\\=\lim\limits_{x\to 1^+}\dfrac{3}{x+1}\\=\lim\limits_{x\to 1^+}\dfrac{3}{1+1}\\=\dfrac{3}{2}$
Since, $f(1)=2$, we can see that our result does not satisfy the statement because $\dfrac{3}{2}\ne2$. Therefore, $f(x)$ is not continuous at $1$.
