A function can be described locally as a graph of a $C^1$ function (Implicit Function Theorem)

Question

Let S be the surface in R3 defined by the equation $z^2y^3+x^2y=2$.

Use the Implicit Function Theorem to determine near which points S can be described locally as the graph of a $C^1$ function $z = f(x, y)$.

In my Calculus book, the following equations from the Implicit Function Theorem are given: $$\frac{\partial z}{\partial x}= -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$ $$\frac{\partial z}{\partial y}= -\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$

I changed the $z^2y^3+x^2y=2$ function to: $$F(x,y,z) = z^2y^3 + x^2y -2 = 0$$

But now do not know what to do and actually I also do not really understand the question. Could someone help me out, please?

Edit later:

So the first order partial derivative w.r.t. z is: $$F_z = 2zy^3$$

And $$F_z\neq 0$$

For this to be true: x can be any number and y and z are not allowed to be $0$.

But how does this answer the question?

Answer 1

In order to write $0=F(x,y,z)=F(x,y,z(x,y))$, you need $F_z \neq 0$. You can see this from your equations for the other partial derivatives, which have zero denominator otherwise. For which points does this occur?

Answer 2

But how does this answer the question?

You just anwswered it in the previous line: $S$ can be described locally as the graph of a function $z=f(x,y)$ near the points of the form $(x,y,z)$, with both $y$ and $z$ different from zero.