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in Hartshorne, Section 4 we have the description of the blow-up of $y^2=x^2(x+1)$ at the origin, that curve have two singularities at $(0,0)$ and $(0,-2/3)$.
But the equations of the blow-up defines the exceptional divisor and a component defined by $y=xu$ and $u^2=x+1$. This component seems to be smooth for me.
why I can not see the singularity corresponding to $(0,-2/3)$ in the blow-up?
It does not sound to me that $(0,-2/3)$ is even on the curve -- if $x=0$ then clearly there is only one $y$ on the curve, namely $y = \pm \sqrt{0(0+1)} = 0$. Similarly, if $y = -2/3$, the equation $x^2(x+1) = 4/9$ defines at most 3 real solutions for $x$, none of which is 0
