Find slope of [(2/x) - f(-3) / x + 3 ] I'm trying to find slope of the following equation, it will then be used to find the equation of the tangent line. 
lim x-> -3 [(2/x) - f(3) / x + 3]
f(x) = (2/x) P(-3, (-2/3))
Use the definition mtan = lim x->a f(x) - f(a) / x - a to find the slope of the line tangent to the graph of f at P
So I already plugged in the points but when solving I eventually get to 
[2 + (2/3)x / x^2 + 3x] and I'm not sure how to get the slope value from this. The answer is supposed to be -2/9

Unlike the title in your posts suggests, you are asked to find the slope of the line tangent to $f(x) = \frac 2x$ at the point $\left(-3, -\frac 23\right)$.
To do this, we take $$\lim_{x\to a}\frac{f(x) - f(a)}{x - a}$$
Given $f(x) = \frac 2x$ and $a = -3$, we have:
$$\lim_{x \to -3}\dfrac{\frac 2x + \left(\frac 23\right)}{x + 3} =  \lim_{x\to -3}\frac{\frac{6 + 2x}{3x}}{x+3} = \lim_{x\to -3} \frac{\frac{2(x+3)}{3x}}{x+3} =\lim_{x \to -3} \frac 2{3x} = -\frac 29$$
