We are given the function $f(x, y) = 3e^{2x} \cos y$. We need to verify that $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$.

1. Problem Description

We are given the function $f(x, y) = 3e^{2x} \cos y$. We need to verify that $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$.

2. Solution Steps

First, we find the first partial derivatives:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (3e^{2x} \cos y) = 3 \cos y \frac{\partial}{\partial x} (e^{2x}) = 3 \cos y (2e^{2x}) = 6e^{2x} \cos y$.

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (3e^{2x} \cos y) = 3e^{2x} \frac{\partial}{\partial y} (\cos y) = 3e^{2x} (-\sin y) = -3e^{2x} \sin y$.

Next, we compute the second partial derivatives:

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial y} (6e^{2x} \cos y) = 6e^{2x} \frac{\partial}{\partial y} (\cos y) = 6e^{2x} (-\sin y) = -6e^{2x} \sin y$.

$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial x} (-3e^{2x} \sin y) = -3 \sin y \frac{\partial}{\partial x} (e^{2x}) = -3 \sin y (2e^{2x}) = -6e^{2x} \sin y$.

Since $\frac{\partial^2 f}{\partial y \partial x} = -6e^{2x} \sin y$ and $\frac{\partial^2 f}{\partial x \partial y} = -6e^{2x} \sin y$, we can conclude that $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$.

3. Final Answer

$\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} = -6e^{2x} \sin y$