 Solution

Let $D=y(3y-2)$. Implicit differentiation of the two constraints gives the linear systemAdding them, then substituting back, gives at regular pointsThe relevant Jacobian determinant is $2(1+a)yz(3y-2)$. Thus the special parameter value is $\boxed{a=-1}$, assuming the other factors are nonzero. At that parameter, adding the original constraints gives $(x+1{)}^2=0$: the intersection curve lies wholly in the plane $x=-1$ and cannot be parametrized by $x$ on an interval. Calling its derivatives “infinite” describes a vertical projection, not a finite-valued differentiable pair $y(x),z(x)$ at $a=-1$; the displayed ratios then have simultaneous zero numerators on the intersection and are undefined.

There are also point-dependent failures when $y=0$, $y=2/3$ or $z=0$. These can produce vertical tangents or require a different local parameter, and simultaneous numerator zeros need separate analysis. The implicit function theorem guarantees finite $y^{\prime },z^{\prime }$ only where the stated Jacobian determinant is nonzero. This distinction is the geometry of a singular projection of an implicit curve.