Question. Given $y^{2}=x^{3}+ax+b$, find the points on the curve where the tangent line is vertical.
Attempt. Let $f(x,y)=x^{3}-y^{2}+ax+b=0$
The tangent is vertical at points where the gradient is horizontal.
The gradient is horizontal if and only if $$\begin{cases} -2y=0\\ 3x^{2}+a\neq0\\ y^{2}=x^{3}+ax+b \end{cases}\Longleftrightarrow\begin{cases} x^{2}\neq-\frac{a}{3}\\ x^{3}+ax+b=0 \end{cases}$$
I'm not sure how to continue forward.