Do all complex zeros in the strip of $\frac{\zeta(s)}{\Gamma(s)} - \frac{\Gamma(1-s)}{\zeta(1-s)}$ lie on the critical line?

Numerical evidence suggests that the complex zeros of:

$$f(s):=\frac{\zeta(s)}{\Gamma(s)} - \frac{\Gamma(1-s)}{\zeta(1-s)}$$

all reside on the line $\Re(s)=\frac12$, except for a finite few outside the critical strip. These zeros come in pairs that each "embrace" a non-trivial zero ($\rho$) of $\zeta(s)$ infinitely tightly. Obviously $f(\rho)$ induces a pole.

The function can rewritten into:

$$f(s):=\frac{\zeta(s) - \dfrac{\pi \csc(\pi \, s)}{\zeta(1-s)}}{\Gamma(s)}$$

so that it will vanish when:

$$\zeta(s)\,\zeta(1-s) = \pi \csc(\pi \, s)$$

When $\Re(s)=\frac12$ both sides will be real and the RHS decreases quickly and monotonically towards zero. Therefore, assuming the RH, each $\rho$ will induce two increasingly nearby 'neighbour' zeros when it pierces through the 'close to zero' RHS-line (see graph below for $\rho_1$). To my surprise I found that $f(s)$ seems to have only one additional zero in the strip at $\frac12 \pm 0.581221780239... i$ that is unrelated to a $\rho$. In the critical strip, $\pi \csc(\pi \, s)$ will always be complex when $\Re(s) \ne \frac12$. For $0 \lt \Re(s) \lt \frac12$ its real part will be always positive and its imaginary part always negative. And vice versa for $\frac12 \lt \Re(s) \lt 1$. Therefore a hypothetical zero of $f(s)$ off the critical line requires both the real and the imaginary parts of $\zeta(s)\zeta(1-s)$ and $\pi \csc(\pi \, s)$ to be equal.

Question:

Since each zero of $f(s)$ (except for the first) is increasingly close to a $\rho$, does the well behaved $\pi \csc(\pi \, s)$ put any constraints on the derivatives of $\zeta(s)\zeta(1-s)$ at a (hypothetical) $\rho$ lying off the critical line?

EDIT: Additional thought (doubt if useful, but I share it anyway):

The following relation must be true at hypothetical zero of $f(s)$ lying off the critical line:

$$\dfrac{\Im\big(\zeta(s)\,\zeta(1-s)\big)}{\Re\big(\zeta(s)\,\zeta(1-s)\big)}=\dfrac{\Im\big(\pi \csc(\pi \, s)\big)}{\Re\big(\pi \csc(\pi \, s)\big)}$$

Using that $\dfrac{\Im\big(\pi \csc(\pi \, s)\big)}{\Re\big(\pi \csc(\pi \, s)\big)}\sim -\cot(\pi \,\Re(s)) =\dfrac{\Re(s)}{\pi}\dfrac{\zeta(1+2\,\Re(s)) \, \zeta(1-2\,\Re(s))}{\zeta(2\,\Re(s)) \, \zeta(-2\,\Re(s))}$ (the latter equation can be derived from just multiplying the reflection formulae of $\zeta(s)$ and $\zeta(-s)$ which induces a $\cos$ divided by a $\sin$ from multiplying the $\Gamma$'s), it follows that at a hypothetical zero this asymptotic relation should hold:

$$\dfrac{\Im\big(\zeta(s)\,\zeta(1-s)\big)}{\Re\big(\zeta(s)\,\zeta(1-s)\big)} \sim \dfrac{\Re(s)}{\pi}\dfrac{\zeta(1+2\,\Re(s)) \, \zeta(1-2\,\Re(s))}{\zeta(2\,\Re(s)) \, \zeta(-2\,\Re(s))}$$