As we know that if $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\neq0$ at $x_0$, then the equation has a horizontal line at $x_0$. Moreover, when $\frac{dx}{dt}=0$ and $\frac{dy}{dt}\neq0$, then there is a vertical line.
What happens if both $\frac{dy}{dt}=0$ and $\frac{dx}{dt}=0$?
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If we have $\frac{dx}{dt} = f(x,y)$ and $\frac{dy}{dt} = g(x,y)$ then we define the term y-null clines where these are the curves(often the lines) on which $\frac{dy}{dt} = 0$. Similarly we define x-null clines. When both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ equal $0$ then the points where this condition is satisfied are called equilibrium points,i.e, where the x-null clines and the y-null clines intersect. It is only at these points is the condition satisfied. Hope it helps.
This happens on the straight diagonal line $(t^3,t^3)$, where nothing of interest happens to the curve, and it happens with $(t^2,t^3)$, which has a sharp corner (a so-called cusp). Also, there is $(t^3,t^5)$, which is nice and smooth and horizontal for $t=0$, and $(t^5,t^3)$, which is vertical.
In short, we need more information than just the first order derivatives in to conclude anything about the shape of the curve in these cases. Higher order derivatives, for instance, could help differentiate between the different cases, but it's possible to construct cases where even that is not enough.