As Dyland Moreland points out $\mathbb{R}^0=\{p\}$. Thus, there is for each discrete space $X$ a unique zero-dimensional manifold structure with the charts $\psi_x:\{x\}\to\mathbb{R}^0$ being the unique such maps.
Let $p\in M$, let $f\in C^\infty\left(N\right)$, and let $X\in T_pM$. By assumption, $\left(F_*X\right)\left(f\right)=X\left(f\circ F\right)=0$. Let $\left(U,\varphi\right)$ be a smooth chart containing $p$. Then
$$X=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_p=\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)},$$
which implies that
$$\left(\sum_iX^i\left(\varphi^{-1}\right)_*\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\right)\left(f\circ F\right)=\sum_iX^i\left.\frac{\partial}{\partial x^i}\right|_{\varphi\left(p\right)}\left(f\circ F\circ\varphi^{-1}\right)=0,$$
which in turn, by basic calculus, implies that $F$ is constant on $U$, i.e., $F\left(x\right)=c$ for some $c\in N$ and every $x\in U$.
Since $M$ is connected, it is the case that $M$ is path connected. Let $q\in M$ and let $\gamma:\left[0,1\right]\to M$ be a path connecting $p$ and $q$. Since, as above, $F$ is constant on each smooth chart $\left(U_{\gamma\left(x\right)},\varphi_{\gamma\left(x\right)}\right)$ containing $\gamma\left(x\right)$ for every $x\in\left[0,1\right]$, it is the case that $F\equiv c$ on $M$ since $F\left(p\right)=c$ and $\gamma$ is continuous.
Best Answer
Hint: With the chain's rule we have $$\frac{\partial(\ \ )}{\partial x}=\frac{\partial(\ \ )}{\partial \rho}\frac{\partial\rho}{\partial x}+\frac{\partial(\ \ )}{\partial\theta}\frac{\partial\theta}{\partial x},$$ i.e. you can see that $$\frac{\partial}{\partial x}= \frac{\partial\rho}{\partial x} \frac{\partial}{\partial \rho}+ \frac{\partial\theta}{\partial x} \frac{\partial}{\partial\theta},$$ as an operator. Similarly $$\frac{\partial}{\partial y}= \frac{\partial\rho}{\partial y} \frac{\partial}{\partial \rho}+ \frac{\partial\theta}{\partial y} \frac{\partial}{\partial\theta},$$
Now, if $$x=\rho\cos\theta,$$ $$y=\rho\sin\theta,$$ you going to get $V=\rho\dfrac{\partial}{\partial\rho}$, taking into account that $$\rho=\sqrt{x^2+y^2},$$ $$\theta=\arctan\frac{y}{x}.$$