= Solution
Put $F=y^2z-x^3+3xz^2-2z^3$. Its partials are
$$
F_x=-3x^2+3z^2,\qquad F_y=2yz,\qquad F_z=y^2+6xz-6z^2.
$$
At infinity $z=0$, the curve equation forces $x=0$, giving $[0:1:0]$, where $F_z\ne0$. In the chart $z=1$, the <derivative> equations give $y=0$ and $x=\pm1$. Only $x=1$ lies on the curve and also makes $F_z=0$. Thus \b[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.
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