Consider a real planar singular cubic curve. After an affine change of coordinates we may assume that the singularity is at the origin $(0,0)$, and that the equation of the curve is $$y^2 - e x^2 + a y^3 + b x y^2 + c x^2 y + d x^3=0$$ where $a$, $b$, $c$, $d \in \mathbb{R}$, not all $0$, $e\in \mathbb{R}$. (Assume also that the cubic is irreducible).
The curve has a rational parametrization $$r(t) =(x(t), y(t)) = \left(\frac{e-t^2}{ a t^3 + b t^2 + c t + d}, \frac{et-t^3}{a t^3 + b t^2 + c t + d}\right)$$
The inflection points of the curve occur at the points for which $r'(t) \parallel r''(t)$. Now, the determinant of the matrix $(r'(t), r''(t))$ equals $$\frac {2( (a e +c) t^3 + 3 (b e+d) t^2 + 3 e(a e +c) t + e(b e+d))}{(a t^3 + b t^2 + c t + d)^3}$$
However, the discriminant of a cubic $p t^3 + 3 q t^2 + 3 e p t + e q$ equals $$-108e(e p^2-q^2)^2$$
Conclusion:
If a real cubic curve has a real node singularity ($e>0$) then it cannot have three real inflection points.
Is there an algebraic geometric reason behind it? I cannot see it right now. Thank you for your interest!
