If $f(x)=\{\begin{array}{ll}0,& x=0\\ x-3,& x>0\end{array}$, then the function $f(x)$ is:

For $x>0$, $f(x)=x-3$.To check if the function is strictly increasing, we find the first derivative:
$$f^{\prime }(x)=\frac{d}{dx}(x-3)=1$$
Since $f^{\prime }(x)=1>0$ for all $x>0$, the function is strictly increasing in the domain $x>0$.At $x=0$, $f(0)=0$, but ${\lim }_{x\to 0^{+}}f(x)=0-3=-3$. Since $f(0)\ne {\lim }_{x\to 0^{+}}f(x)$, the function is not continuous at $x=0$. However, the question asks specifically about the behavior for $x>0$.