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Write $F=xyz+yzw+zwx+wxy$. A projective hypersurface singularity must have $F=0$ and all four first partial derivatives zero. If all coordinates were nonzero, writeOn $F=0$, differentiation gives $F_x=-xyzw/x^2\ne 0$, a contradiction. Thus some coordinate is zero; by symmetry take $w=0$. The surface equation gives $xyz=0$, so a second coordinate is zero, say $z=0$. Then $F_z=xy$, so a third coordinate is zero. The remaining four projective possibilities indeed have every first derivative zero. Therefore all singular points areOver the usual characteristic-zero ground field, each is an ordinary double point: near $[1:0:0:0]$, the affine equation has nondegenerate quadratic part $yz+zw+wy$, and symmetry treats the others.