A radial or "central" force field $\mathbf{F}$ in the plane can be written in the form $\mathbf {F}(x,y)=f(r)\,\mathbf{r} $, where $\mathbf {r}=x\mathbf{i}+y\mathbf {j}$ and $r=\Vert \mathbf {r}\Vert$. Show that such a force field is conservative on $\Bbb R^2 - \{0\}$.
It is can be easily shown that it is a gradient on $\Bbb R^2-\{(0,0)\}$ as shown by @JackyChong here (Prove that a radial force field is conservative). I am guessing that $\mathbf{F}$ is also a gradient on $\Bbb R^2$ which is what I'm unable to prove.
The main issue as one can guess is with the point $(0,0)$ as $\nabla \mathbf{r}$ doesn't exist at $(0,0)$. My idea was to rewrite the line integral, $\int_{a}^{b} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt$ as
$$\int_{a}^{c-\epsilon }\textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt\ \ +\ \ \int_{c-\epsilon}^{ c+\epsilon} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt\ \ +\ \ \int_{c+\epsilon}^{ b} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt$$
by lineaity of line integrals.
Now assuming that $r(\alpha(c))=0 \ ie \ \alpha(c)=0$ and $\forall x \in [a,b]-c\ ,\ \alpha(x)\ne0$
Then we have $$\int_{a}^{b} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt = \phi(c-\epsilon)-\phi(a)+\int_{c-\epsilon}^{ c+\epsilon} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt-\phi(c-\epsilon)_+\phi(b)$$
where $\phi(r) = \int^r_0 \tau f(\tau)\ d\tau$.
Let $\varphi(\epsilon) = \int_{c-\epsilon}^{ c+\epsilon} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt$
Now using the fact that $\varphi(\epsilon)$ and $\phi(r)$ are left continuous at zero, hence
$$\int_{a}^{b} \textbf{F}(\alpha(t))\cdot \boldsymbol{\alpha}(t)^{'} dt\ \ =-\phi(a)+\phi(b)$$
Now to truly show $\textbf{F}$ is a gradient we need this for all piece-wise smooth paths and these paths can pass through $0$ arbitrary number of times which I'm unable to deal with using the simply minded idea above.
You're working too hard. Referring to the answer in the linked post, you can see directly that if you set $\phi(x,y) = F(\sqrt{x^2+y^2})$, then (using the actual definition of the partial derivatives at the origin) $$\nabla\phi(0,0) = (0,0) = \mathbf F(0,0).$$ Thus, you have a potential function on all of $\Bbb R^2$, not just away from the origin.