Answer
$$\nabla g =\left\langle \frac{-x^2+y^2}{\left(x^2+y^2\right)^2} ,-\frac{2xy}{\left(x^2+y^2\right)^2} \right\rangle$$
Work Step by Step
Given
$$ g(x, y)=\frac{x}{x^{2}+y^{2}}$$
Since
\begin{align*}
\frac{\partial g}{\partial x}&=\frac{\frac{\partial \:}{\partial \:x}\left(x\right)\left(x^2+y^2\right)-\frac{\partial \:}{\partial \:x}\left(x^2+y^2\right)x}{\left(x^2+y^2\right)^2}= \frac{-x^2+y^2}{\left(x^2+y^2\right)^2} \\
\frac{\partial g}{\partial y}&=x\frac{\partial \:}{\partial \:y}\left(\frac{1}{x^2+y^2}\right)= -\frac{2xy}{\left(x^2+y^2\right)^2}
\end{align*}
Then
\begin{align*}
\nabla g&=\left\langle\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle\\
&=\left\langle \frac{-x^2+y^2}{\left(x^2+y^2\right)^2} ,-\frac{2xy}{\left(x^2+y^2\right)^2} \right\rangle
\end{align*}
