Answer
$$\frac{\partial z}{\partial x}=\frac{y^3-x^2y}{(x^2+y^2)^2}$$
$$\frac{\partial z}{\partial y}=\frac{x^3-y^2x}{(x^2+y^2)^2}$$
Work Step by Step
To solve both partial derivatives we will use quotient rule.
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{x^2+y^2}\right)=\frac{(x^2+y^2)\frac{\partial }{\partial x}(xy)-xy\frac{\partial }{\partial x}(x^2+y^2)}{(x^2+y^2)^2}=\frac{(x^2+y^2)\cdot y-xy\cdot2x}{(x^2+y^2)^2}=\frac{x^2y+y^3-2x^2y}{(x^2+y^2)^2}=\frac{y^3-x^2y}{(x^2+y^2)^2}$$
because $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\frac{xy}{x^2+y^2}\right)=\frac{(x^2+y^2)\frac{\partial}{\partial y}(xy)-xy\frac{\partial }{\partial y}(x^2+y^2)}{(x^2+y^2)^2}=\frac{(x^2+y^2)\cdot x-xy\cdot2y}{(x^2+y^2)^2}=\frac{x^3+y^2x-2y^2x}{(x^2+y^2)^2}=\frac{x^3-y^2x}{(x^2+y^2)^2}$$
because now $x$ is treated as a constant.