We know that $\phi_y=-1$, $\phi_z=0$ thanks to the Implicit Function Theorem. Moreover $\phi \in C^1$ in an environment of $(1,0)$, so $\lim_{t\to 1} \phi(t,t^3-1)=1$. Hence,
$$\lim_{t\to1} \frac{\phi(t,t^3-1)-t}{t^2-1}$$
is an indeterminate form (i.e. "$\frac{0}{0}$"). Since both numerator and denominator are differentiable, we can apply L'Hôpital's Rule.
$$\lim_{t\to1} \frac{\phi(t,t^3-1)-t}{t^2-1} \overset{\text{provided it exists}}= \lim_{t\to1} \frac{\phi'(t,t^3-1)}{2t}$$
But what the heck is $\phi'(t,t^3-1)$?
Unless I'm missing something, we have $\phi(y,z) = \sqrt{2-y^2-z^3}$ on a neighbourhood of $(1,0)$ so
$$\lim_{t\to1} \frac{\phi(t,t^3-1)-t}{t^2-1} = \lim_{t\to 1} \frac{\sqrt{2-t^2-(t^3-1)^3}-t}{t^2-1} \stackrel{\text{L'H}}{=} \lim_{t\to 1}\frac{\frac{-2t-9t^2(t^3-1)}{2\sqrt{2-t^2-(t^3-1)^3}}-1}{2t} = -1$$