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 $C=A+Bx\in\mathbb Q(A,B)$.

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