Answer
$$\frac{\partial g(x,y)}{\partial x}=\frac{x}{x^2+y^2}$$
$$\frac{\partial g(x,y)}{\partial y}=\frac{y}{x^2+y^2}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial g(x,y)}{\partial x}=\frac{\partial }{\partial x}(\ln\sqrt{x^2+y^2})=\frac{1}{\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(\sqrt{x^2+y^2})=\frac{1}{\sqrt{x^2+y^2}}\cdot\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(x^2+y^2)=\frac{1}{2(x^2+y^2)}\cdot2x=\frac{x}{x^2+y^2}$$
Here $y$ is treated as a constant. Also, we used chain rule to find partial derivatives $\frac{\partial }{\partial x}(\ln\sqrt{x^2+y^2})$ and $\frac{\partial }{\partial x}(\sqrt{x^2+y^2}).$
The partial derivative with respect to $y$ is:
$$\frac{\partial g(x,y)}{\partial y}=\frac{\partial }{\partial y}(\ln\sqrt{x^2+y^2})=\frac{1}{\sqrt{x^2+y^2}}\frac{\partial }{\partial y}(\sqrt{x^2+y^2})=\frac{1}{\sqrt{x^2+y^2}}\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial y}(x^2+y^2)=\frac{1}{2(x^2+y^2)}\cdot2y=\frac{y}{x^2+y^2}$$
Now is $x$ treated as a constant. Again, we used chain rule to find partial derivatives $\frac{\partial }{\partial y}(\ln\sqrt{x^2+y^2})$ and $\frac{\partial }{\partial y}(\sqrt{x^2+y^2}).$
