= Solution
Put $q=\varphi'(y-\beta z)$. Apply <implicit differentiation> with respect to $x$ and $y$, using the <chain rule>:
$$
1-\alpha z_x=-\beta qz_x,\qquad -\alpha z_y=q(1-\beta z_y).
$$
Consequently
$$
(\alpha-\beta q)z_x=1,\qquad (\alpha-\beta q)z_y=-q.
$$
At a <point> where the assumed <differentiable function> $z(x,y)$ exists, the first equation forces $\alpha-\beta q\ne0$. Combining the equations therefore gives
$$
\boxed{\alpha z_x+\beta z_y=\frac{\alpha-\beta q}{\alpha-\beta q}=1.}
$$
The assertion is understood on such differentiable branches of the <implicit function>; arbitrary parameter choices need not produce one everywhere.
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