Answer
Not continuous.
Work Step by Step
$f(x)$ will be continuous at $x=0$ if $f(0)=\lim_{x\to 0}f(x)$. We know that $f(0)=1$.
$$\lim_{x\to 0}f(x)\\=\lim_{x\to 0}\frac{x^3+3x}{x^2-3x}\\=\lim_{x\to 0}\frac{x(x^2+3)}{x(x-3)}\\=\lim_{x\to 0}\frac{(x^2+3)}{(x-3)}\\=\frac{(0^2+3)}{(0-3)}=-1$$
$1\ne-1$, hence it is not continuous at $0$.