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Solution
Your input: find $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{2} y^{2}\right)$$$
First, find $$$\frac{\partial}{\partial y}\left(x^{2} y^{2}\right)$$$
Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=x^{2}$$$ and $$$f=y^{2}$$$:
$${\color{red}{\frac{\partial}{\partial y}\left(x^{2} y^{2}\right)}}={\color{red}{x^{2} \frac{\partial}{\partial y}\left(y^{2}\right)}}$$Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=2$$$:
$$x^{2} {\color{red}{\frac{\partial}{\partial y}\left(y^{2}\right)}}=x^{2} {\color{red}{\left(2 y^{-1 + 2}\right)}}=2 x^{2} y$$Thus, $$$\frac{\partial}{\partial y}\left(x^{2} y^{2}\right)=2 x^{2} y$$$
Next, $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{2} y^{2}\right)=\frac{\partial}{\partial y} \left(\frac{\partial}{\partial y}\left(x^{2} y^{2}\right) \right)=\frac{\partial}{\partial y}\left(2 x^{2} y\right)$$$
Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=2 x^{2}$$$ and $$$f=y$$$:
$${\color{red}{\frac{\partial}{\partial y}\left(2 x^{2} y\right)}}={\color{red}{2 x^{2} \frac{\partial}{\partial y}\left(y\right)}}$$Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial y} \left(y \right)=1$$$:
$$2 x^{2} {\color{red}{\frac{\partial}{\partial y}\left(y\right)}}=2 x^{2} {\color{red}{1}}$$Thus, $$$\frac{\partial}{\partial y}\left(2 x^{2} y\right)=2 x^{2}$$$
Therefore, $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{2} y^{2}\right)=2 x^{2}$$$
Answer: $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{2} y^{2}\right)=2 x^{2}$$$