Given f(x) = (1)/(sqrt(x+3)), we have f(x+h) = (1)/(sqrt(x+h+3)).

The derivative of a function $f(x)$ is given by:

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

Given $f(x) = \frac{1}{\sqrt{x+3}}$, we have $f(x+h) = \frac{1}{\sqrt{x+h+3}}$.
Substitute these into the formula:

$$f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h+3}} - \frac{1}{\sqrt{x+3}}}{h}$$

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

The conjugate of $\sqrt{x+3} - \sqrt{x+h+3}$ is $\sqrt{x+3} + \sqrt{x+h+3}$.

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

The numerator becomes:

$$(\sqrt{x+3})^2 - (\sqrt{x+h+3})^2 = (x+3) - (x+h+3) = x+3 - x - h - 3 = -h$$

Substitute this back into the expression:

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

$$f'(x) = \frac{-1}{\sqrt{x+0+3}\sqrt{x+3}(\sqrt{x+3} + \sqrt{x+0+3})}$$ $$f'(x) = \frac{-1}{\sqrt{x+3}\sqrt{x+3}(\sqrt{x+3} + \sqrt{x+3})}$$ $$f'(x) = \frac{-1}{(x+3)(2\sqrt{x+3})}$$

Since $\sqrt{x+3} = (x+3)^{1/2}$, we can write $(x+3)\sqrt{x+3}$ as $(x+3)^1 (x+3)^{1/2} = (x+3)^{1 + 1/2} = (x+3)^{3/2}$.

$$f'(x) = \frac{-1}{2(x+3)^{3/2}}$$

The derivative of $y = \frac{1}{\sqrt{x+3}}$ using the first principle is:

$$-\frac{1}{2(x+3)^{3/2}}$$