let $ F: {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}, (x_0, y_0, z_0)^{T}=(2,2,2)^{T}$ and $F(x,y,z)=x^3-2xyz+3xz^2+3z^3$
I first had to find $(x_0, y_0, z_0)^{T}$ such that $F(x_0, y_0, z_0)=40$, I came to the solution of $x_0=y_0=z_0=2$
now
Prove that in a neighborhood T of $(x_0, y_0)^{T}$ there exists an implicit function $h(x,y)$, such that $F(x,y,h(x,y))=40$ $\forall (x,y) \in T$
I calculated all the partial derivatives of $F$ and showed that they are continuous in $(x_0, y_0)^{T}$, but I do not know how to continue from here? is $h(x,y) =2$?
To apply the implicit function theorem here, the last thing you should verify is that the partial derivative of $F$ with respect to $z$ at the particular solution you found (i.e. at the point $(2,2,2)$) is not zero. If that is verified, then the implicit function theorem assures you that such a function $h$ exists, but in general there isn't a way to give the function $h$ explicitly.