The vector function is considered being smooth if the curve is continuous and the differential $\dfrac{\mathrm dr}{\mathrm dt}$ is not $0$ at any point.
I can understand the first condition saying that the curve must be continuous but cannot understand how the differential $\dfrac{\mathrm dr}{\mathrm dt}$ equals to $0$ makes the curve not smooth?
Will you help prove the notion for me?
Here's a picture of the curve parameterized by $t\mapsto (t^3,t^2)$:
Notice how this has a clearly visible point where its direction changes abruptly (a cusp). This is why one can have problems when the derivative goes to zero - basically, if the parameterization stops moving at some point, then it "forgets" about what came before, and can go off in whatever direction it pleases, creating corners and cusps. Thus, one has the requirement that smooth curves have a parameterization where the derivative does not go to zero anywhere.