**Solution:**

To find the derivative of the given function Y=(x^2-1)(2x-1)/(x-3), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative of this function is given by:

(dy/dx) = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2

In this case, f(x) = (x^2-1)(2x-1) and g(x) = (x-3). Let's compute the derivatives of f(x) and g(x) separately before substituting them into the quotient rule formula.

**Finding the derivative of f(x):**

To find the derivative of f(x) = (x^2-1)(2x-1), we can use the product rule. The product rule states that if we have a function of the form f(x)*g(x), then the derivative of this function is given by:

(f(x)*g(x))' = f'(x)*g(x) + f(x)*g'(x)

Let's find the derivatives of x^2-1 and 2x-1 separately.

f'(x) = (d/dx)(x^2-1) = 2x
g'(x) = (d/dx)(2x-1) = 2

Now we can substitute these derivatives into the product rule formula:

f'(x)*g(x) = 2x*(x-3)
f(x)*g'(x) = (x^2-1)*2

**Finding the derivative of g(x):**

To find the derivative of g(x) = (x-3), we can use the power rule. The power rule states that if we have a function of the form f(x) = (x^n), then the derivative of this function is given by:

(d/dx)(x^n) = n*x^(n-1)

In this case, n = 1. So, the derivative of g(x) = (x-3) is:

g'(x) = (d/dx)(x-3) = 1

**Applying the quotient rule:**

Now that we have the derivatives of f(x) and g(x), we can substitute them into the quotient rule formula:

(dy/dx) = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2

(dy/dx) = ((x-3)*2x - (x^2-1)*2)/((x-3))^2

Simplifying the expression:

(dy/dx) = (2x^2 - 6x - 2x^2 + 2)/((x-3))^2

(dy/dx) = (-6x + 2)/((x-3))^2

Therefore, the derivative of Y=(x^2-1)(2x-1)/(x-3) is (dy/dx) = (-6x + 2)/((x-3))^2.