Start with $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$, then using the cyclic property of the trace, its linearity, and the expansion of the logarithm of an operator, you get $$ \text{Tr}\ln(\mathbb{I}_4\ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln((\gamma^5)^2 \ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln(\gamma^5(i\gamma^\mu\partial_\mu-m)\gamma^5) $$ Then using the fact that $\{\gamma^5, \gamma^\mu\} = 0$, the $\gamma^5$'s act on the $\gamma^\mu$ by conjugation to yield $-\gamma^\mu$, while $m$ commutes with the $\gamma^5$, yielding $$ \text{Tr}\ln(-i\gamma^\mu\partial_\mu-m) $$

So, using the linearity of the trace again, we can break up $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ into $\frac12 \text{Tr} [ \ln(i\gamma^\mu\partial_\mu-m) + \ln(i\gamma^\mu\partial_\mu-m)]$ and rewrite the second one using the previous identity.

The crux of the argument is proving that $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ = $\text{Tr}\ln(-i\gamma^\mu\partial_\mu-m)$. While it might be possible to prove this using the fact that the trace of the log of an operator is the logarithm of the product of its eigenvalues, I wasn't able to do this myself.

Start with $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$, then using the cyclic property of the trace, its linearity, and the expansion of the logarithm of an operator, you get $$ \text{Tr}\ln(\mathbb{I}_4\ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln((\gamma^5)^2 \ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln(\gamma^5(i\gamma^\mu\partial_\mu-m)\gamma^5) $$ Then using the fact that $\{\gamma^5, \gamma^\mu\} = 0$, the $\gamma^5$'s act on the $\gamma^\mu$ by conjugation to yield $-\gamma^\mu$, while $m$ commutes with the $\gamma^5$, yielding $$ \text{Tr}\ln(-i\gamma^\mu\partial_\mu-m) $$

So, using the linearity of the trace again, we can break up $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ into $\frac12 \text{Tr} [ \ln(i\gamma^\mu\partial_\mu-m) + \ln(i\gamma^\mu\partial_\mu-m)]$ and rewrite the second one using the previous identity.

Start with $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$, then using the cyclic property of the trace, its linearity, and the expansion of the logarithm of an operator, you get $$ \text{Tr}\ln(I_4\ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln((\gamma^5)^2 \ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln(\gamma^5(i\gamma^\mu\partial_\mu-m)\gamma^5) $$ Then using the fact that $\{\gamma^5, \gamma^\mu\} = 0$, the $\gamma^5$'s act on the $\gamma^\mu$ by conjugation to yield $-\gamma^\mu$, while $m$ commutes with the $\gamma^5$, yielding $$ \text{Tr}\ln(-i\gamma^\mu\partial_\mu-m) $$

So, using the linearity of the trace again, we can break up $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ into $\frac12 \text{Tr} [ \ln(i\gamma^\mu\partial_\mu-m) + \ln(i\gamma^\mu\partial_\mu-m)]$ and rewrite the second one using the previous identity.

Start with $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$, then using the cyclic property of the trace, its linearity, and the expansion of the logarithm of an operator, you get $$ \text{Tr}\ln(\mathbb{I}_4\ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln((\gamma^5)^2 \ (i\gamma^\mu\partial_\mu-m)) \\ = \text{Tr}\ln(\gamma^5(i\gamma^\mu\partial_\mu-m)\gamma^5) $$ Then using the fact that $\{\gamma^5, \gamma^\mu\} = 0$, the $\gamma^5$'s act on the $\gamma^\mu$ by conjugation to yield $-\gamma^\mu$, while $m$ commutes with the $\gamma^5$, yielding $$ \text{Tr}\ln(-i\gamma^\mu\partial_\mu-m) $$

So, using the linearity of the trace again, we can break up $\text{Tr}\ln(i\gamma^\mu\partial_\mu-m)$ into $\frac12 \text{Tr} [ \ln(i\gamma^\mu\partial_\mu-m) + \ln(i\gamma^\mu\partial_\mu-m)]$ and rewrite the second one using the previous identity.