= Solution
Write
$$
f_1=xy,
\qquad
f_2=y^2-z^3+xz,
\qquad
f_3=x(x+y+2z+1).
$$
The equation $xy=0$ splits the analysis into two cases. If $x=0$, the remaining equation is $y^2=z^3$, giving the <cuspidal cubic>
$$
C=V(x,y^2-z^3).
$$
It is irreducible because $y^2-z^3$ is irreducible in $k[y,z]$: as a quadratic in $y$, it could factor only if $z^3$ were a square in $k[z]$.
If $y=0$, the other equations become
$$
z(x-z^2)=0,
\qquad
x(x+2z+1)=0.
$$
Their solutions are $(0,0,0)$, which already lies on $C$, and the two points
$$
P=(-1,0,0),
\qquad
Q=(1,0,-1).
$$
Thus the <irreducible components> are
$$
\boxed{C,\quad\{P\},\quad\{Q\}}.
$$
Equivalently, the <radical ideal> of the <cusp with two isolated-point components> is
$$
\boxed{I(V)
=(x,y^2-z^3)
\cap(x+1,y,z)
\cap(x-1,y,z+1).}
$$
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