Given f(x) = x2 + (3)/(x).

Given $f(x) = x^2 + \frac{3}{x}$.
Then $f(x+h) = (x+h)^2 + \frac{3}{x+h}$.
Expand $(x+h)^2$:

$$f(x+h) = x^2 + 2xh + h^2 + \frac{3}{x+h}$$

$$f(x+h) - f(x) = \left(x^2 + 2xh + h^2 + \frac{3}{x+h}\right) - \left(x^2 + \frac{3}{x}\right)$$ $$f(x+h) - f(x) = x^2 + 2xh + h^2 + \frac{3}{x+h} - x^2 - \frac{3}{x}$$ $$f(x+h) - f(x) = 2xh + h^2 + \frac{3}{x+h} - \frac{3}{x}$$

Combine the fractional terms:

$$f(x+h) - f(x) = 2xh + h^2 + \frac{3x - 3(x+h)}{x(x+h)}$$ $$f(x+h) - f(x) = 2xh + h^2 + \frac{3x - 3x - 3h}{x(x+h)}$$ $$f(x+h) - f(x) = 2xh + h^2 - \frac{3h}{x(x+h)}$$

$$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - \frac{3h}{x(x+h)}}{h}$$

Factor out $h$ from the numerator:

$$\frac{f(x+h) - f(x)}{h} = \frac{h(2x + h) - \frac{3h}{x(x+h)}}{h}$$ $$\frac{f(x+h) - f(x)}{h} = 2x + h - \frac{3}{x(x+h)}$$

$$f'(x) = \lim_{h \to 0} \left(2x + h - \frac{3}{x(x+h)}\right)$$

Substitute $h=0$ into the expression:

$$f'(x) = 2x + 0 - \frac{3}{x(x+0)}$$ $$f'(x) = 2x - \frac{3}{x^2}$$

The derivative of $x^2 + \frac{3}{x}$ from first principles is $\boxed{2x - \frac{3}{x^2}}$.