Answer
$$\frac{\partial z}{\partial x}=-2x\sin(x^2+y^2)$$
$$\frac{\partial z}{\partial y}--2y\sin(x^2+y^2)$$
Work Step by Step
We will use chain rule to find both partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\cos(x^2+y^2))=-\sin(x^2+y^2)\frac{\partial }{\partial x}(x^2+y^2)=-\sin(x^2+y^2)\cdot2x=-2x\sin(x^2+y^2)$$
We treated $y$ as a constant here.
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\cos(x^2+y^2))=-\sin(x^2+y^2)\frac{\partial }{\partial y}(x^2+y^2)=-\sin (x^2+y^2)\cdot2y=-2y\sin(x^2+y^2)$$
because now $x$ is treated as a constant.