Find $y^{\prime}$ by (a) applying the Product Rule and (b) multiplying the factors to produce a sum of simpler terms to differentiate. \[ y=\left(2 x^{2}+3\right)\left(8 x+5+\frac{1}{x}\right) \]

Step 1 ：First, we recognize that the given function can be written in the form \((a+b)(c+d+e)\), where \(a=2x^2\), \(b=3\), \(c=8x\), \(d=5\), and \(e=\frac{1}{x}\).

Step 2 ：Next, we apply the Product Rule for differentiation, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Step 3 ：Applying the Product Rule, we get \(y^\prime = (2x^2+3)\prime(8x+5+\frac{1}{x})+(2x^2+3)(8x+5+\frac{1}{x})\prime\).

Step 4 ：The derivative of \(2x^2+3\) is \(4x\), and the derivative of \(8x+5+\frac{1}{x}\) is \(8-\frac{1}{x^2}\).

Step 5 ：Substituting these derivatives into the equation, we get \(y^\prime = (4x)(8x+5+\frac{1}{x})+(2x^2+3)(8-\frac{1}{x^2})\).

Step 6 ：Multiplying through, we get \(y^\prime = 32x^2+20x+\frac{4x}{x}+16x^2-6+\frac{3}{x^2}\).

Step 7 ：Simplifying, we get \(y^\prime = 48x^2+20x-6+\frac{3}{x^2}\).

Step 8 ：Finally, we can rewrite the derivative in a more standard form as \(y^\prime = 48x^2+20x-6+3x^{-2}\).

Step 9 ：\(\boxed{y^\prime = 48x^2+20x-6+3x^{-2}}\) is the final answer.