Answer
$$\frac{\partial^2z}{\partial x^2}=\frac{y^2}{\sqrt{x^2+y^2}(x^2+y^2)},\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial^2z}{\partial y\partial x}=-\frac{xy}{\sqrt{x^2+y^2}(x^2+y^2)},\frac{\partial^2z}{\partial y^2}=\frac{x^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
Work Step by Step
The partial derivative with respect to $x$ is:
$$\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}(\sqrt{x^2+y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2x=\frac{x}{\sqrt{x^2+y^2}}$$
The partial derivative with respect to $y$ is:
$$\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}(\sqrt{x^2+y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\partial y}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2y=\frac{y}{\sqrt{x^2+y^2}}$$
The second partial derivatives are:
$$\frac{\partial ^2z}{\partial x^2}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial x}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big)=\frac{\frac{\partial}{\partial x}(x)\sqrt{x^2+y^2}-x\frac{\partial}{\partial x}(\sqrt{x^2+y^2})}{(\sqrt{x^2+y^2})^2}=\frac{\sqrt{x^2+y^2}-x\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(x^2+y^2)}{x^2+y^2}=\frac{x^2+y^2-x^2}{(x^2+y^2)\sqrt{x^2+y^2}}=\frac{y^2}{(x^2+y^2)\sqrt{x^2+y^2}}$$
$$\frac{\partial ^2}{\partial y^2}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial y}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)=\frac{\frac{\partial}{\partial y}(y)\sqrt{x^2+y^2}-y\frac{\partial}{\partial y}\sqrt{x^2+y^2}}{(\sqrt{x^2+y^2})^2}=\frac{\sqrt{x^2+y^2}-y\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\partial y}(x^2+y^2)}{x^2+y^2}=\frac{x^2+y^2-y^2}{(x^2+y^2)\sqrt{x^2+y^2}}=\frac{x^2}{(x^2+y^2)\sqrt{x^2+y^2}}$$
$$\frac{\partial^2z}{\partial y\partial x}=\frac{\partial}{\partial y}\Big(\frac{\partial z}{\partial x}\Big)=\frac{\partial}{\partial y}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big)=-x\frac{1}{2(\sqrt{x^2+y^2})^3}\frac{\partial}{\partial y}(x^2+y^2)=-\frac{xy}{(\sqrt{x^2+y^2})^3}$$
$$\frac{\partial ^2z}{\partial x\partial y}=\frac{\partial}{\partial x}\Big(\frac{\partial z}{\partial y}\Big)=\frac{\partial }{\partial x}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)=-y\frac{1}{2(\sqrt{x^2+y^2})^3}\frac{\partial}{\partial x}(x^2+y^2)=-\frac{xy}{(\sqrt{x^2+y^2})^3}$$
We see that the second mixed partial derivatives are equal.
