Answer
$\nabla \phi (x, y)=\lt \dfrac{y}{y^2+x^2}, \dfrac{-x}{y^2+x^2}\gt $
Work Step by Step
Our aim is to compute the the gradient vector field vector.
Here, we have $\phi (x, y)=\tan^{-1} (x/y)$
Now, $\nabla \phi (x, y)=\lt \dfrac{\partial \phi (x, y)}{\partial x} , \dfrac{\partial \phi (x, y)}{\partial y} \gt $
or, $\nabla \phi (x, y)=\lt \dfrac{\partial (\tan^{-1} (x/y))}{\partial x} , \dfrac{\partial (\tan^{-1} (x/y))}{\partial y} \gt \\= \lt \dfrac{1}{1+(x/y)^2} \dfrac{\partial (x/y)}{\partial x} , \dfrac{1}{1+(x/y)^2} \dfrac{\partial (x/y)}{\partial y} \gt\\=\lt \dfrac{1}{y} \times \dfrac{y^2}{y^2+x^2}, \dfrac{-x}{y^2} \times \dfrac{y^2}{y^2+x^2}\gt $
Thus, we have $\nabla \phi (x, y)=\lt \dfrac{y}{y^2+x^2}, \dfrac{-x}{y^2+x^2}\gt $
