20. Find the four second partial derivatives of the following function.
\[
\begin{array}{l}
h(x, y)=x^{4}+x y^{2}+5 \\
\frac{\partial^{2} h}{\partial x^{2}}=\square \\
\frac{\partial^{2} h}{\partial y^{2}}=\square \\
\frac{\partial^{2} h}{\partial y \partial x}=\square \\
\frac{\partial^{2} h}{\partial x \partial y}=\square
\end{array}
\]

Step 1 ：Given the function \(h(x, y) = x^4 + xy^2 + 5\)

Step 2 ：Find the first partial derivatives: \(\frac{\partial h}{\partial x} = 4x^3 + y^2\) and \(\frac{\partial h}{\partial y} = 2xy\)

Step 3 ：Find the second partial derivatives: \(\frac{\partial^2 h}{\partial x^2} = \frac{\partial}{\partial x}(4x^3 + y^2)\), \(\frac{\partial^2 h}{\partial y^2} = \frac{\partial}{\partial y}(2xy)\), \(\frac{\partial^2 h}{\partial y \partial x} = \frac{\partial}{\partial y}(4x^3 + y^2)\), and \(\frac{\partial^2 h}{\partial x \partial y} = \frac{\partial}{\partial x}(2xy)\)

Step 4 ：Calculate these derivatives using the power rule and the chain rule to get: \(\frac{\partial^2 h}{\partial x^2} = 12x^2\), \(\frac{\partial^2 h}{\partial y^2} = 2x\), \(\frac{\partial^2 h}{\partial y \partial x} = 2y\), and \(\frac{\partial^2 h}{\partial x \partial y} = 2y\)

Step 5 ：Final Answer: \(\boxed{\frac{\partial^{2} h}{\partial x^{2}}=12x^{2}}\), \(\boxed{\frac{\partial^{2} h}{\partial y^{2}}=2x}\), \(\boxed{\frac{\partial^{2} h}{\partial y \partial x}=2y}\), \(\boxed{\frac{\partial^{2} h}{\partial x \partial y}=2y}\)