Tangent line at $(0,0)$ of plane curve $y^3=x^5$ Let $f(x,y)=y^3-x^5$. $\nabla f(x,y)=(-5x^4,3y^2)$. $\nabla f(0,0)=(0,0)$ so $(0,0)$ is a singular point of the curve $y^3-x^5=0$. 
On the other hand, it makes sense to define $g(x)=x^{5/3}$ for $x \in \mathbb{R}$. Then the curve $y^3-x^5=0$ is the graph of the function $g$. But $g'(x)=\frac{5}{3} x^{2/3}$ so $g'(0)=0$, so the graph of $g$ has the tangent line $y=0$ at the point $(0,0)$. 
Therefore, there is something I am not understanding by what it means to say that $(0,0)$ is a singular point of the curve $y^3-x^5=0$, because there seems to be a tangent line at this point. So either being a singular point does not imply not having a tangent line, or I am somehow being sloppy in the definition of tangent line. 
I would dearly like to clear this up. I am not a differential or algebraic geometer but an analyst; I've been thinking about this for teaching a multivariable calculus course I've been teaching.

In algebraic geometry, for a  curve defined by an equation $P(x, y)=0$, where $P\in K[X,Y]\;$ ($K$ a field), the tangents at origin have equation the homogeneous  part of lowest degree  in the equation. In the present case, it means the tangents are given by $\;y^3=0$, so it is the line $y=0$, counting for three.
