= Solution
On a smooth implicit branch, put $Q=x+z^2$ and require $Q\ne0$. This is exactly the <implicit function theorem> condition, since the defining equation has vertical <partial derivative> $3Q$. Differentiating the cubic relation in $x$ and $y$ gives
$$
Qz_x=-z,\qquad Qz_y=1,
\qquad z_x=-\frac zQ,\quad z_y=\frac1Q.
$$
For a second <derivative>, the dependence of $Q$ on $z$ must also be included: $Q_x=1+2zz_x$ and $Q_y=2zz_y$. Thus
$$
\begin{aligned}
z_{xx}&=-\frac{z_x}{Q}+\frac{z(1+2zz_x)}{Q^2}=\frac{2xz}{Q^3},\\
z_{yy}&=-\frac{2zz_y}{Q^2}=-\frac{2z}{Q^3},\\
z_{xy}&=-\frac{z_y}{Q}+\frac{2z^2z_y}{Q^2}=\frac{z^2-x}{Q^3}.
\end{aligned}
$$
The required cancellation follows immediately:
$$
\boxed{z_{xx}+xz_{yy}=\frac{2xz}{(x+z^2)^3}-\frac{2xz}{(x+z^2)^3}=0.}
$$
The verification is local to smooth branches with $x+z^2\ne0$; the implicit relation does not guarantee a differentiable function at its singular points.
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