Missing $1/r$ factor in gradient of polar function
Polar coordinates:
$$
\begin{cases} x = r\cos(\theta) \\ y = r \sin(\theta) \end{cases} \quad \Longrightarrow \quad
\begin{cases} \mathbf {\hat r} = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} \\
\mathbf {\hat \theta} = \begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \end{bmatrix} \end{cases}
$$
Chain rules:
$$
\frac{\partial f}{\partial r} =
\frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}
= \cos(\theta) \frac{\partial f}{\partial x} + \sin(\theta) \frac{\partial f}{\partial y} \\
\frac{\partial f}{\partial \theta} =
\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}
= - r \sin(\theta) \frac{\partial f}{\partial x} + r \cos(\theta) \frac{\partial f}{\partial y}
$$
Matrix format:
$$
\begin{bmatrix} \Large \frac{\partial f}{\partial r} \\ \Large \frac{1}{r} \frac{\partial f}{\partial \theta} \end{bmatrix} =
\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ - \sin(\theta) & \cos(\theta) \end{bmatrix}
\begin{bmatrix} \Large \frac{\partial f}{\partial x} \\ \Large \frac{\partial f}{\partial y} \end{bmatrix}
$$
Inverse transform:
$$
\begin{bmatrix} \Large \frac{\partial f}{\partial x} \\ \Large \frac{\partial f}{\partial y} \end{bmatrix} =
\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}
\begin{bmatrix} \Large \frac{\partial f}{\partial r} \\ \Large \frac{1}{r} \frac{\partial f}{\partial \theta} \end{bmatrix}
= \frac{\partial f}{\partial r} \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix}
+ \frac{1}{r} \frac{\partial f}{\partial \theta} \begin{bmatrix} - \sin(\theta) \\ \cos(\theta) \end{bmatrix}
$$
Conclusion:
$$
\nabla f = \frac{\partial f}{\partial r}\mathbf {\hat r} + \frac 1r\frac{\partial f}{\partial \theta}\mathbf {\hat \theta}
$$