To differentiate the given expression X=(A+C^3)/(B-C^2/3) and then logartihmize it, we first need to find the derivative of X with respect to C.

Let Y = (A+C^3)/(B-C^2/3)

To differentiate Y with respect to C, we need to use the quotient rule:

dY/dC = (B-C^2/3)(d/dC)(A+C^3) - (A+C^3)(d/dC)(B-C^2/3) / (B-C^2/3)^2

dY/dC = (B-C^2/3) (3C^2) - (A+C^3) (-2C/3) / (B-C^2/3)^2

dY/dC = (3BC^2 - C^4) - (-2AC - 2C^4)/3 / (B-C^2/3)^2

dY/dC = (3BC^2 - C^4 + 2AC + 2C^4)/3 / (B-C^2/3)^2

dY/dC = (3BC^2 + 2AC - C^4)/3 / (B-C^2/3)^2

Now, let's take the natural logarithm of Y:

ln(Y) = ln((3BC^2 + 2AC - C^4)/3 / (B-C^2/3)^2)

ln(Y) = ln(3BC^2 + 2AC - C^4) - ln((B-C^2/3)^2)

Therefore, the differentiated and natural logarithmized expression is:

dY/dC = (3BC^2 + 2AC - C^4)/3 / (B-C^2/3)^2
ln(Y) = ln(3BC^2 + 2AC - C^4) - ln((B-C^2/3)^2)