When is a curve smooth at points where $dy/dx$ does not exist? The curve $y=x^{1/3}$ is smooth everywhere even though $dy/dx$ does not exist at $x=0$. Why?
In general; Wherever $dy/dx$ does not exist on a curve, how can I show that it could still be smooth at those places?
 A: The issue here is that if you have a continuously differentiable function $f(x,y)$, its level set $f(x,y)=c$ will be smooth curves whenever one of $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ is nonzero at each point of the curve. 
Your curve is the graph $y=x^{1/3}$, and this can be rewritten as the $0$-level curve of $f(x,y)=y^3-x$ (this remark is equivalent to @Pratyush Sarkar's answer). Note that another example of your phenomenon is the curve $f(x,y)=x^2+y^2-1$. You might start with $y=\sqrt{1-x^2}$ and $dy/dx$ doesn't exist at $x=\pm 1$, and yet the circle is certainly a smooth curve; here we have $\dfrac{\partial f}{\partial x}(\pm1,0) \ne 0$, even though $\dfrac{\partial f}{\partial y}(\pm1,0) = 0$. Recall that the formula for implicit differentiation tells us that
$$\text{when } \dfrac{\partial f}{\partial y} \ne 0, \text{   we have } \frac{dy}{dx} = -\frac{\partial f/\partial x}{\partial f/\partial y}\,.$$
A: The curve $\gamma_1(x) = (x, x^{\frac{1}{3}})$ can be reparametrized as $\gamma_2(y) = (y^3, y)$. The second curve $\gamma_2$ is differentiable everywhere with ${\gamma_2}'(y) \neq (0, 0)$ for all $y$. So the curve is smooth.
