Define $F(x,y) := x^3 - y^2$ and $C$ is the curve $F(x,y) = 0$.
Now when you graph this, we can see its clearly not $C^1$, as there is a cusp at the origin (meaning not differentiable at origin).
We can define $x$ globally as a function of $y$, but according to Implicit Function Theorem, $F_x(0,0) = 0$, so we can't locally define $x$ as a function of $y$. The question is essentially asking if this is a contradiction of the Implicit Function Theorem and I believe it's not since $x^3 - y^2 = 0$ is not $C^1$, so the assumptions of IFT are not met.
The partials are $(3x^2, -2y)$ which exist and are continuous everywhere, which imply that it is $C^1$, but this is clearly not the case.
How do I go about showing the curve $x^3-y^2=0$ is not $C^1$?
There are several issues with this. First of all $F$ is indeed a $C^1$ function as a polynomial in two variables. However it's total derivative at $(x,y)=0$ is $0$. More importantly $0$ is not a regular value of $F$. This means we can not use the implicit function theorem since it requires full range of the differential map. When you define the curve $x$ as a function of $y$ you will not get a differentiable function at $y=0$. This is however not enough to show that the curve isn't $C^1$ since it could potentially be defined by a different function and this is a deeper question but it doesn't matter. Since the IFT doesn't describe equivalence but only an implication, so even if $C$ is a smooth curve, it could be such irrespective of $F$.