Finding slope m of tangent to curve been years since I took calculus and am currently struggling with how to properly work out the following: 

Using the tanget line slope formula: 

My understanding is that I would need to find the slope with the following approach: 

Now, at this point, don't I want to rationalize the numerator? Or the denominator? 
If I go with rationalizing the numerator (multiplying by the conjugate), I do the following, but end up with a denominator that looks really overly complicated:
 
What am I doing wrong here? 
 A: I have got $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{3}{x}-\frac{3}{a}}{x-a}=\frac{3(a-x)}{ax(x-a}=-\frac{3}{ax}$$
It is $$\frac{\frac{3}{\sqrt{x}}-\frac{3}{\sqrt{a}}}{x-a}=\frac{3(\sqrt{a}-\sqrt{x})(\sqrt{a}+\sqrt{x})}{\sqrt{ax}(x-a)(\sqrt{a}+\sqrt{x})}=-\frac{3}{\sqrt{ax}}$$
A: You are darn close.  Rationalizing the numerator is the way to go.
$\frac {3(\sqrt a - \sqrt x)(\sqrt a + \sqrt x)}{\sqrt{ax}(x-a)(\sqrt a + \sqrt x)}$
I factored out a 3, so that I wouldn't have to cancel it later on.
Rather than multiplying out the denominator.. just leave it alone for a moment.
$\frac {3(a-x)}{\sqrt{ax}(x-a)(\sqrt a + \sqrt x)}$
We have a factor in numerator and denominator that we can factor.
$\frac {-3}{\sqrt{ax}(\sqrt a + \sqrt x)}$
Now as we let $x$ approach $a.$
$\frac {-3}{\sqrt{a^2}(\sqrt a + \sqrt a)} = -\frac {3}{2a^\frac 32}$
A: What about observing that
$a - x = (\sqrt a + \sqrt x)(\sqrt a - \sqrt x)? \tag 1$
whence
$\dfrac{3 \sqrt a - 3\sqrt x}{\sqrt{xa} (x - a)} = \dfrac{3(\sqrt a - \sqrt x)}{\sqrt{xa}(\sqrt x + \sqrt a)(\sqrt x - \sqrt a)} = -\dfrac{3}{\sqrt{ax}(\sqrt a + \sqrt x)}; \tag 2$
now as $x \to a$ we have
$\displaystyle \lim_{x \to a} \dfrac{3 \sqrt a - 3\sqrt x}{\sqrt{xa} (x - a)}= -\dfrac{3}{\sqrt{a^2}(\sqrt a + \sqrt a)} = -\dfrac{3}{2\sqrt{a^2}{\sqrt a}} = -\dfrac{3}{2(\sqrt a)^3} = -\dfrac{3}{2}x^{-3/2}. \tag 3$
