so if the PDF attains its maximum at 0 it means that $f'(0) = 0$:
$$f'(x) = -\frac{x}{\sigma^2 \sqrt{\pi}}\exp({-\frac{(x - \mu)^2}{2\sigma^2}}) $$
$$f'(0) = 0 \iff 0 =0$$
yep, no valuable information whatsoever,
now onto the second derivative criterion : $f''(0) < 0$
$$f''(0) = 0 -\frac{1}{\sigma^2 \sqrt{\pi}}\exp({-\frac{\mu^2}{2\sigma^2}}) < 0$$
again, no valuable information whatsoever.
maybe if we try using the fact that $f(0) \geq f(y), \, \forall y \in \mathbb{R}$ for $y = \mu$
then :
$$\frac{1}{\sigma\sqrt{2 \pi}}\exp(-\frac{\mu^2}{2\sigma^2}) \geq\frac{1}{\sigma\sqrt{2 \pi}}\iff \exp(-\frac{\mu^2}{2\sigma^2}) \geq1 \iff -\frac{\mu^2}{2\sigma^2} \geq 0 \iff \mu^2 \leq 0$$
therefore $\mu = 0$ since $0$ is the only 'negative' positive number.
I'm pretty confident that my work is correct but I'd like a confirmation, thanks !