The function given is:

$f(x) = \frac{2}{x} + 3x^2 - 3, \quad x \neq 0.$

To find $f'(x)$, we need to differentiate each term of $f(x)$ with respect to $x$.

Step-by-Step Solution

1. Differentiate $\frac{2}{x}$: $\frac{d}{dx} \left( \frac{2}{x} \right) = \frac{d}{dx} \left( 2x^{-1} \right) = -2x^{-2} = -\frac{2}{x^2}.$

2. Differentiate $3x^2$: $\frac{d}{dx} (3x^2) = 6x.$

3. Differentiate the constant term $-3$: $\frac{d}{dx} (-3) = 0.$

Combine Results

Now, we combine these results to obtain $f'(x)$: $f'(x) = -\frac{2}{x^2} + 6x.$

Final Answer

$f'(x) = -\frac{2}{x^2} + 6x.$

Would you like further details or have any questions?

Additional questions for deeper understanding:

1. How would $f'(x)$ change if $f(x)$ included a term with $x^3$?
2. What does $f'(x) = 0$ imply about the function $f(x)$ at certain points?
3. How would the derivative change if the constant term $-3$ were different?
4. Could we interpret $f'(x)$ geometrically? What does it tell us about the slope of $f(x)$?
5. How does the differentiation of $\frac{2}{x}$ change if we consider higher powers, like $\frac{2}{x^n}$?

Tip: Always rewrite terms with negative exponents or fractions before differentiating; it simplifies the process.