If you use the $\log$ property for expressions of the form $\log \left(\frac{m}{n}\right)$ and then take the derivative of $f(x)=\ln \left(\frac{e^{x}-e^{3 x}}{e^{-2 x}+e^{-3 x}}\right)$, you obtain...
A. $\frac{-5 e^{x}+4}{e^{x}-1}$
B. $\frac{-5 e^{x}+4}{-e^{x}-1}$
C. $\frac{e^{x}-e^{3 x}(3)}{e^{x}-e^{3 x}}+\frac{e^{-2 x}(-2)+e^{-3 x}(-3)}{e^{-2 x}+e^{-3 x}}$
D. $\frac{e^{x}-e^{3 x}(3)}{e^{x}-e^{3 x}}-\frac{e^{-2 x}(-2)+e^{-3 x}(-3)}{e^{-2 x}+e^{-3 x}}$
E. $\frac{-2}{e^{x}-e^{3 x}}-\frac{5}{e^{-2 x}+e^{-3 x}}$
F. $-\frac{-5 e^{x}+4}{e^{x-1}}$

Step 1 ：We are given the function $f(x)=\ln \left(\frac{e^{x}-e^{3 x}}{e^{-2 x}+e^{-3 x}}\right)$ and we are asked to find its derivative.

Step 2 ：We can use the quotient rule for derivatives and the chain rule to solve this problem. The quotient rule is $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ and the chain rule is $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$.

Step 3 ：First, we need to find the derivatives of the numerator and the denominator. The derivative of $e^{x}-e^{3 x}$ is $e^{x}-3e^{3 x}$ and the derivative of $e^{-2 x}+e^{-3 x}$ is $-2e^{-2 x}-3e^{-3 x}$.

Step 4 ：Then, we can apply the quotient rule. The derivative of the function $f(x)$ is $\frac{(-3e^{3x} + e^{x})(e^{-2x} + e^{-3x}) - (-e^{3x} + e^{x})(-2e^{-2x} -3e^{-3x})}{(e^{-2x} + e^{-3x})^2}$.

Step 5 ：This expression is a bit complicated, but we can simplify it to get the final answer.

Step 6 ：The simplified derivative of the function $f(x)$ is $\frac{-5 e^{x}+4}{e^{x}-1}$.

Step 7 ：So, the final answer is \(\boxed{\frac{-5 e^{x}+4}{e^{x}-1}}\).