Calculate the derivative using limit definition. This is the function $f(x)$$=\frac{1}{\sqrt{3x-2}}$ .
I wrote that $$\lim_{h\to 0}\frac{\frac{\sqrt{3x+3h-2}}{3x+3h-2}-\frac{\sqrt{3x-2}}{3x-2}}{h}.$$
I am not able to continue further.

Let $a = \sqrt{3x+h-2}$ and $b = \sqrt{3x-2}$, then $a^2 - b^2 = h$ which means $h \to 0$ as $a \to b$. So we've:
$$ \lim_{a \to b} \frac{1/a-1/b}{a^2-b^2}  = \lim_{a \to b} \frac{b-a}{(ab)(a+b)(a-b)} = - \lim_{a \to b} \frac{1}{ab(a+b)} = -\frac{1}{2 b^3} = -\frac{1}{2\sqrt{(3x-2)^3}.} $$
