= Solution
Let $D=y(3y-2)$. <Implicit differentiation> of the two constraints gives the <linear system>
$$
D y'-2zz'=-2x,\qquad aD y'+2zz'=-2.
$$
Adding them, then substituting back, gives at regular points
$$
\boxed{\frac{dy}{dx}=-\frac{2(x+1)}{(1+a)y(3y-2)},\qquad
\frac{dz}{dx}=\frac{ax-1}{(1+a)z}.}
$$
The 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 $y',z'$ only where the stated <Jacobian determinant> is nonzero. This distinction is the geometry of a <singular projection of an implicit curve>.
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