I am only posting this as an answer because it annoys me to see a question like this listed as "unanswered", thus "hovering" near the top of the list of unanswered questions. If dhy wants to write up his comment as an answer, then I will delete this answer.

The surface given by the OP is as far as possible from being "general type". Just to remind, a projective variety $S$ over a characteristic $0$ field is of **general type** if one, and hence any, smooth, projective model of the variety has big dualizing sheaf. This is a birational property. In particular, since the dualizing sheaf on $\mathbb{P}^n$, namely $\mathcal{O}_{\mathbb{P}^n}(-n-1)$, is anti-ample, rational varieties are not of general type. As dhy correctly points out, much more is known: for a diagram $$\begin{array}{lcc} T & \xrightarrow{f} & S \\ \downarrow \pi \\ R \end{array},$$ such that $f$ is dominant and generically finite and such that $\pi$ is projective and flat of positive fiber dimension, then the geometric generic fiber of $\pi$ is of general type (this can easily fail in positive characteristic). In particular, if there exists a diagram as above such that the geometric generic fiber of $\pi$ is rational or an Abelian variety, then also $S$ is not of general type. (Once upon a time, Serge Lang had a conjecture that was roughly the converse of this statement, but nobody seems to believe that conjecture anymore.)

Back to the surface $S$ of the OP, i.e., the zero locus in
$\mathbb{A}^3$ of $$F(x,y,z) = z^6 + x^4 - y^2,$$ this is an irreducible, affine hypersurface with an isolated singularity at the origin. By the way, such a singularity is called *quasi-homogeneous*, in this case with weights $(3,6,2)$. This came up a little while ago, because for a quasi-homogeneous function $$g:\mathbb{A}^r \to \mathbb{A}^1,$$ the critical locus of $g$ is $\{0\}\subset \mathbb{A}^1$. Anyway, the open subset $U= D(xz)\cap S$ of $S$ is dense. Since being of general type is birational, it suffices to work with $U$. Denote $w = x/z\in \mathcal{O}^{\times}_{\mathbb{A}^3}(D(xz))$. Then the defining equation of $S$ becomes, $$ \widetilde{F}(w,x,y) = x^4(w^{-6}x^2+1) - y^2.$$ Since $x$ is invertible on $D(xz)$, also $y/x^2$ is in $\mathcal{O}_{\mathbb{A}^3}(D(xz))$. Thus the defining equation is the same as, $$F(w,x,y) = x^4\left( 1 - \left(\frac{y}{x^2}\right)^2 + (w^{-3}x)^2 \right).$$ Now denote $$ v = \frac{y}{x^2} + \frac{x}{w^3}.$$ Then there is an isomorphism of $k$-algegras $$\frac{k[x,y,z][1/xz]}{\langle F(x,y,z) \rangle} \xrightarrow{\cong} k[v,w]\left[\frac{1}{vw}\right],$$ $$x = \frac{w^3(v-v^{-1})}{2}, \ y = \frac{w^6(v-v^{-1})^2(v+v^{-1})}{8}, \ z = \frac{w^2(v-v^{-1})}{2}.$$ In particular, $U$ is isomorphic to $\mathbb{G}_m^2$. Therefore $S$ is a rational surface. So $S$ is not of general type.

One commenter points out that for general rational functions $a(z)$, $b(z)$ and $c(z)$, for the element $$ F(x,y,z) = x^4 +a(z)x^2 + b(z)x + c(z) - y^2,$$ of $k(z)[x,y]$, the zero locus is an elliptic surface, hence not of general type.

`KodairaEnriquesType(S : CheckADE := true);`

$\endgroup$ – joro Aug 16 '15 at 9:053more comments