To differentiate the function \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x, we can use the quotient rule.

The quotient rule states that if we have a function \(f(x) = \frac{g(x)}{h(x)}\), then \(f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}\).

In this case, we have \(g(x) = 6x^3 - 5x^2 + 1\) and \(h(x) = 3x^2\).

Taking the derivative of g(x) with respect to x, we get \(g'(x) = 18x^2 - 10x\).

Taking the derivative of h(x) with respect to x, we get \(h'(x) = 6x\).

Substituting these values into the quotient rule formula, we get \(f'(x) = \frac{(18x^2 - 10x)(3x^2) - (6x^3 - 5x^2 + 1)(6x)}{(3x^2)^2}\).

Simplifying this expression, we get

\(f'(x) = \frac{54x^4 - 30x^3 - 36x^4 + 30x^3 - 6x}{9x^4} = \frac{18x^4 - 6x}{9x^4} = 2 - \frac{2}{3x^3}\).

So, the derivative of \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x is \(2 - \frac{2}{3x^3}\).