How do you see that the cubic surface given by $WXZ+Y^2Z+X^3=0$ has an $A_5$ singularity at $[W:X:Y:Z]=[1:0:0:0]$? (Reference: Class XIX https://www.singsurf.org/parade/Cubics.php)
I was thinking one could either resolve it or compute the colength of the ideal generated by the partials (milnor number), but I wondered if there was a way that provided more insight (similar to how an $A_n$ curve singularity can be detected using curvilinear schemes or a way to put it into the normal form $x^2+y^2+z^6$ for an $A_5$ surface singularity).
Projection from this singular point gives a rational map from the cubic surface to $\mathbb{P}^2$. Its inverse is the map $$ \mathbb{P}^2 \dashrightarrow \mathbb{P}^3,\qquad (x:y:z) \mapsto \left(- \frac{y^2z + x^3}{xz} : x : y : z \right). $$ Thus, it blows up the subscheme of $\mathbb{P}^2$ defined by equations $$ xz = y^2z + x^3 = 0 $$ and then takes the anticanonical image of the blowup.
It is easy to trace that there will be six (-2) curves on the blowup: one (out of two) exceptional divisor over the point $(0:0:1)$ and a chain of length 5 consisting of strict transforms of lines $x = 0$ and $z = 0$ and three (out of four) exceptional divisors over the point $(0:1:0)$. The first will give an $A_1$ singularity (at point $(0:0:0:1)$) and the second will give the $A_5$ singularity at point $(1:0:0:0)$.