The condition for a singular point is that the Jacobian should have less than full rank. The full rank of a 1×2 matrix is 1. So at a singular point, the matrix must have rank 0, which is the same as saying that both entries should be 0.

Therefore you need to solve the equations

An efficient way to do that is to add the equations, getting . So either , leading to the solutions ; or , leading to the solutions and also the solution .

(You then have to check whether any of those points actually lie on the curve. If they don't, they obviously don't yield singular points.)