I have to prove that $$\vec{F}=\frac{\vec{r}}{r^2}$$ is a conservative vector field without finding it's potential function.
However, $\vec{F}$ has a hole in its domain, at $(x,y,z)=(0,0,0)$, so I can't just check it's curl.
How do I go about proving this?
For this example it still suffices to show that $\operatorname{curl} \vec F = 0$. It's true that the domain of $\vec F$ has a hole, but the domain is still simply connected: Any loop in the domain, say, based at a point can be continuously deformed to the constant path that sits at that point.
Alternatively, in spherical coordinates $(r, \theta, \varphi)$, the gradient is $$\operatorname{grad} f = f_r \vec r + \frac{1}{r} f_{\theta} \vec \theta + \frac{1}{r \sin \theta} f_{\varphi} \vec \varphi.$$ In particular, the gradient of a radial function $f(r)$ is $$\operatorname{grad} (f(r)) = f_r(r) \vec r ,$$ so the Fundamental Theorem of Calculus guarantees that any continuous radial vector field (not defined at the origin, where spherical coordinates behave badly) is conservative.