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The function is not hypertranscendental. Indeed, let $A=e^x$ and $B=e^{e^x}$. Then we have $x'=1,A'=A$ and $B'=AB$. These equalities imply that the field $\mathbb Q(x,A,B)$ is closed under differentiation. Since this field has transcendence degree (at most) $3$ over $\mathbb Q$, we see that for any $C\in\mathbb Q(A,B)$, the elements $C,C',C'',C'''$ must be algebraically dependent, which implies $C$ is not hypertranscendental. Now we can just take $C=A+Bx\in\mathbb Q(A,B)$.