Consider the lower branch of the hyperbola $f(x)=\frac{1}{x}.$ We can add up the images on the curve corresponding to pre-images equal to $-{n}$ for $n=1,2,3,\cdot\cdot\cdot.$
Upon doing this we get, $h(x)=-\sum_{n\ge1} n^x.$ It can be shown with some algebra that $h(x)$ can be expressed as a product of primes.
Changing the coordinate system of $\Bbb R^2$ via diffeomorphism, $g:\Bbb R^2\to \Bbb R^2$ with $g(x,y)=(e^x,e^y),$ we can define an analogous hyperbola in the new coordinate system, defined by $v(x)=e^{\frac{1}{\log(x)}}.$
If we decide to add up the images (like before) on $v(x)$ corresponding to pre-images equal to $e^{-\frac{1}{n}}$ for $n=1,2,3,\cdot\cdot\cdot$ then we get $\rho(x)=\sum_{n\ge1} e^{-n^x}.$
Why is the Euler product destroyed for $\rho(x)?$ Is $\rho(x)$ connected to primes?
The Euler product "works" because it establishes a 1-1 correspondence between the terms $1+\frac{1}{2^x}+\frac{1}{3^x}+\cdots$ with terms from expanding $\prod_p (1-1/p^x)^{-1}$. When you write $e^{-n^x}$, that correspondence is seemingly lost because exponents add when multiplied. For example replacing it with $\prod_p(1-1/e^{p^x})^{-1}$ doesn't work because you get terms like $\exp(p_1^{x}+p_2^{x}+\cdots)$.