Example 1. Find the second-order partial derivatives of the function $z=\left(x^{2}+y^{2}\right)^{2}$.
Solution. First, we find the first-order partial derivatives:

$$
\frac{\partial z}{\partial x}=2\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}\right)_{x}^{\prime}=2\left(x^{2}+y^{2}\right) 2 x=4 x^{3}+4 x y^{2}
$$

$$
\frac{\partial z}{\partial y}=2\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}\right)_{y}^{\prime}=2\left(x^{2}+y^{2}\right) 2 y=4 x^{2} y+4 y^{3}
$$

Differentiating each of the obtained functions with respect to $x$ and $y$, we get the second-order partial derivatives:

$$
\begin{gathered}
\frac{\partial^{2} z}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}\left(4 x^{3}+4 x y^{2}\right)=\left(4 x^{3}+4 x y^{2}\right)_{x}^{\prime}=12 x^{2}+4 y^{2} \\
\frac{\partial^{2} z}{\partial y^{2}}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\left(4 x^{2} y+4 y^{3}\right)_{y}^{\prime}=4 x^{2}+12 y^{2} \\
\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\left(4 x^{3}+4 x y^{2}\right)_{y}^{\prime}=8 x y \\
\frac{\partial^{2} z}{\partial y \partial x}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\left(4 x^{2} y+4 y^{3}\right)_{x}^{\prime}=8 x y
\end{gathered}
$$

As expected,

$$
\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial^{2} z}{\partial y \partial x}
$$