 Solution

Put $F=y^{2}z-x^3+3x{z}^2-2z^3$. Its partials areAt infinity $z=0$, the curve equation forces $x=0$, giving $[0:1:0]$, where $F_z\ne 0$. In the chart $z=1$, the derivative equations give $y=0$ and $x=\pm 1$. Only $x=1$ lies on the curve and also makes $F_z=0$. Thus the only singular point is $[1:0:1]$. As a check, the affine equation factors as $y^2=(x-1{)}^2(x+2)$, whose quadratic tangent cone there is $y^2-3(x-1{)}^2$, giving two distinct tangent lines.