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Hola Johana sM!
$$\begin{align}&tan^{-1}x=arctanx\\&z=(x^2+y^2)arctan \frac y x\\&\\&D_x=2xarctan \frac y x +(x^2+y^2)· \frac{ \frac {-y}{ x^2}}{1+ \frac{y^2}{x^2}}=2xarctan \frac y x-y\\&\\&D_y=2y arctan \frac y x+(x^2+y^2)· \frac{ \frac 1 x}{1+ \frac{y^2}{x^2}}=2yarctan \frac y x+x\\&\\&D_{yx}=2x \frac{ \frac 1 x}{1+ \frac{y^2}{x^2}}-1=\frac{2x^2}{x^2+y^2}-1= \frac{x^2-y^2}{x^2+y^2}\\&\\&D_{xy}=2y \frac{- \frac y {x^2}}{1+ \frac{y^2}{x^2}}+1=\frac{-2y^2}{x^2+y^2}+1= \frac{x^2-y^2}{x^2+y^2}\\&\\&D_{yy}=2 arctan \frac y x+2y \frac{ \frac 1 x}{1+\frac{y^2}{x^2}}=2 arctan \frac y x+ \frac {2yx}{x^2+y^2}\\&\\&D_{xx}=2 arctan \frac y x+2x \frac{- \frac y {x^2}}{1+ \frac{Y^2}{x^2}}=2 arctan \frac y x- \frac {2xy}{x^2+y^2}\\&\\&\end{align}$$
Saludos
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