An exercise in Silverman This is self-learning, not homework.
Problem: Let $A, B \in \bar{\mathbb{K}}$. Characterize the values of $A, B$ for which each of the following varieties is singular. In particular, as $(A,B)$ ranges over $\mathbb{A}^2$, the "singular values" lie on a one-dimensional subset of $\mathbb{A}^2$, so "most" values of $(A,B)$ give a non-singular variety.
($\mathbb{K}$ is a field, $\mathbb{A}^2$ is affine 2-space, etc.)
$(a) V: Y^2Z + AXYZ + BYZ^2 = X^3$.
$(b) V: Y^2Z = X^3 + AXZ^2 + BZ^3$ (char $\mathbb{K} \neq 2$).
My attempt:
We need to find $(A,B)$ so that the polies defining these varieties have derivatives (with respect to each variable) which have roots in $\mathbb{K}$, i.e. we need to solve the systems of equations
$(a) 2YZ + AXZ + BZ^2 = Y^2 + AXY + 2BYZ = 3 X^2 - AYZ = 0$
$(b) 2YZ = 3X^2 + AZ^2 = 2AXZ + 3BZ^2 - Y^2 = 0$
for $A$ and $B$.
Silverman gives the solutions $(a) B(A^3 - 27B) = 0$ and $(b) 4A^3 + 27B^2 = 0$. This is probably just embarrassingly basic highschool algebra but I'm not seeing it.
 A: Hope you don't mind if I do (b) and leave (a) to you.
$$2YZ=3X^2+AZ^2=2AXZ+3BZ^2-Y^2=0$$
So, we start with just the leftmost equation: $2YZ=0\implies Y=0$ or $Z=0$.
Note that $Z=0\implies X=0\implies Y=0$ from the rest of the equations, and this cannot happen in projective space, so we rule that out.
Now say $Y=0$, then right most equation gives:
$$Z(2AX+3BZ)=0$$
Arguing as before, we rule out $Z=0$, so $2AX+3BZ=0$.
Now assuming $A\neq0$, we have $$X=\frac{3BZ}{2A},$$
substituting that into the last one and noting that $Z\neq0$, we have the following $$3\left(\frac{3BZ}{2A}\right)^2+AZ^2=0\implies \left(\frac{27B^2}{4A^2}+A\right)Z^2=0$$
Hence, result.
A: Thanks, it's clear now. Here's part (a).
$$2YZ + AXZ + BZ^2 = Y^2 + AXY + 2BYZ = 3X^2 - AYZ = 0$$
By similar reasoning to (b), if any of $X,Y,Z$ is $0$, so are the other two. Let's pick $Z = 1$, then the first two equations become
$$2Y + AX + B = Y(Y + AX + 2B) = 0$$
But $Y \neq 0$, so $$Y + AX + 2B = 0$$
On the one hand, adding the two up and using the third equation, we get something nice and suggestive:
$$2AX = -3(B + Y) \\ 3X^2 = AY \implies \frac{27}{4A^2}(B + Y)^2 = AY$$
which gives the quadratic
$$ Y^2 + (2B - \frac{4A^3}{27})Y + B^2 = 0$$
On the other hand, subtracting the two, a miracle occurs:
$$ 2Y + AX + B - (Y + AX + 2B) = Y - B = 0 \implies Y = B$$
So it seems that scaling $Z = 1$ was equivalent to scaling $Y = B$. Putting them together, $$B^2 + (2B - \frac{4A^3}{27})B + B^2 = 4B^2 - \frac{4A^3}{27}B = 0 $$ or $$B(A^3 - 27B) = 0$$
