Partial differentiation dz/ds when function of z is differentiable function of x and y but not given explicitly

Question

$$ z = f(x,y) $$ and $z$ is differentiable.

Let $$ x = s^2 - t, y = t^3 \ln(1+s) $$

Then $$ \frac{\partial z}{\partial s}$$ at $s = 0$ and $t = 0$ is?

What assumption do I have to make? $$ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}(2s) + \frac{\partial z}{\partial y}\left(\frac{1}{1+s}\right) $$

Because I need to know the partial of $z$ with respect to $x$ and the partial of $z$ with respect to $y$.