Given that $ r = \sqrt{x^2 + y^2},\qquad \theta = \arctan\left(\frac{y}{x}\right) $ and $U = U(x,y)$, using the chain rule in two variables, we have $ \frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x} $ $ \frac{\partial U}{\partial y} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial y} $ Now, $ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2} = \frac{x}{\sqrt{x^2 + y^2}} = \cos \theta $ $ \frac{\partial r}{\partial y} = \frac{\partial}{\partial y} \sqrt{x^2+y^2} = \frac{y}{\sqrt{x^2 + y^2}} = \sin \theta $ and $ \frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x} \arctan \left(\frac{y}{x}\right) = -\frac{y}{x^2 + y^2} = -\frac{\sin \theta}{r} $ $ \frac{\partial \theta}{\partial y} = \frac{\partial}{\partial y} \arctan \left(\frac{y}{x}\right) = \frac{y}{x^2 + y^2} = \frac{\cos \theta}{r} $ hence $ \nabla U = U_r \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} + \frac{1}{r} U_\theta \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} $ Lastly, the unitary vector in the radial direction $\hat{r}$ is $ \hat{\textbf{r}} = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} $ and due orthogonality $ \hat{\theta} = \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} $ which leads to the desired form $ \nabla U = U_r \hat{\textbf{r}} + \frac{1}{r} U_\theta \hat{\theta} $