I need help on this calculus homework question. How would I go about solving this with a step by step explanation, all help appreciated! 
http://tinypic.com/r/xckr9z/8 

Find the instantaneous rate of change of the given function when $x=a$. 
  $$H(x) = \frac {\color{red}{6}} x + x; a= 1$$

EDIT:
So for this I have to to use the formula ${f(x+h)-f(x)\over h}$ in order to get the instantaneous rate of change. My original function is $f(x)={6\over x}+x$.
So, I did ${{6\over (x+h)}+(x+h)-({6\over x}+x) \over h}$
Since $x=1$ I ended up with : ${({6\over (1+0)})+(1+0)-(({6\over 1})+1)\over h}$
And ended up with $6+1-6-1$ which ends to $0\over 0$. And that wasn't the answer so I am really unsure n what to do.
NOTE: By the dislikes I realize I went about asking my question in the wrong manner, and I apologize. Hopefully it's correct now, thanks to all the help!
 A: The instantaneous rate of change of the function $H$ at $x=a$ is the derivative of $H$ w.r.t. $x$ at that value of $x$.
You simply need to evaluate: $\left.\dfrac{\mathrm{d}H(x)}{\mathrm{d} x}\right|_{x=a}$
So, find the derivative of: $\left(\dfrac{6}{x} + x\right)$ w.r.t. $x$ and evaluate at $(x=a)$ when $(a=1)$.

You seem to have gotten this mostly right; you just need to simplify and cancel all $h$ terms common to the numerator and denominator before you take the limit $h\to 0$.
$$\begin{align}H'(x) & = \frac{\mathrm{d}H(x)}{\mathrm{d} x} \\ & = \lim\limits_{h\to 0} \frac{H(x+h)-H(x)}{h} \\ & = \lim\limits_{h\to 0} \frac{\left(\frac{6}{x+h}+x+h\right)-\left(\frac 6 x + x\right)}{h} & \text{since }H(x)=\frac 6 x + x \\ & = \lim\limits_{h\to 0} \frac{\left(\frac {6}{x+h} - \frac 6 x\right) + (x + h - x)}{h} & \text{by rearrangement} \\ & = \lim\limits_{h\to 0} \frac{\frac {6x - 6(x+h)}{x(x+h)} + h}{h} & \text{cross multiplying fractions} \\ & = \lim\limits_{h\to 0} \frac{\frac {-6h}{x^2+hx} + h}{h} \\ & = \lim\limits_{h\to 0} \frac {-6}{x^2+hx} + 1 & \color{red}{\text{Cancel common terms.}} \\ & = \frac {-6}{x^2} + 1 & \text{taking to the limit} \\ \therefore H'(1) & = -5 \end{align}$$
