Conditions for a smooth parametric curve A curve defined by $x=f(t), y=g(t)$ is smooth if $f'(x)$ and $g'(x)$ are continuous and not simultaneously zero.
Why do we have the second condition(simultaneously zero)?

That second condition is so that the parametrisation is regular. Informally speaking, you want the curve to be "drawn" by the parametrisation in a nice way. In particular, you don't want the parametrisation to turn back on itself. If you have a regular parametrisation then you know that the only way that $f'(t) = g'(t) = 0$ is when the curve itself is singular.
Consider two examples. Take the cusp $\gamma(t) = (t^2,t^3)$; this has a regular parametrisation. If you take the derivative $\gamma'(t) = (2t,3t^2)$ and so the singular points come from $2t = 3t^2 = 0$, i.e. $t=0$. So the singular point of the curve is $\gamma(0) = (0,0)$, i.e. the cusp point.
Now consider the smooth curve $y=x^2$ with the non-regular parametrisation $\alpha(t) =  (t^3,t^6)$. We have $\alpha'(t) = (3t^2,6t^5)$. It's tempting to think that $t=0$ gives a singular point, but it doesn't. The curve is smooth. It's the parametrisation that causes the problem. Informally, the pen stops for an instant at the origin.
