Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem 
$$f(x)= \frac{x}{x+2}   $$
How do I go about properly solving this, I seemed to get  

$$\frac{x}{x+2}\   $$
  as my answer again? What are the steps I should follow?

I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative. 

First. The derivative function of $f(x)$ is not $\frac{f(x+h)-f(x)}{h}$. (That would be a function of two variables, $x$ and $h$). Rather, the derivative function of $f(x)$ is
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
The limit is important. Without the limit, it's just plain wrong.
Second. I don't know how you "got back" $f(x)$. The computations aren't that bad, just lots of algebra; perhaps you substituted incorrectly?
$$\begin{align*}
f(x) &= \frac{x}{x+2}\\
f(x+h) &= \frac{(x+h)}{(x+h)+2} = \frac{x+h}{x+h+2}\\
f(x+h) - f(x) &= \frac{x+h}{x+h+2} - \frac{x}{x+2}\\
&= \frac{(x+h)(x+2)}{(x+h+2)(x+2)} - \frac{x(x+h+2)}{(x+2)(x+h+2)}\qquad\text{(common denominator)}\\
&= \frac{(x+h)(x+2) - x(x+h+2)}{(x+2)(x+h+2)}\\
&= \frac{x(x+2) + h(x+2) - x(x+2) - xh}{(x+2)(x+h+2)}\\
&= \frac{h(x+2-x)}{(x+2)(x+h+2)}\\
&= \frac{2h}{(x+2)(x+h+2)}.
\end{align*}$$
Therefore,
$$\begin{align*}
f'(x) &= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0}\frac{\quad\frac{2h}{(x+2)(x+h+2)}\quad}{h}\\
&= \lim_{h\to 0}\frac{2h}{h(x+2)(x+h+2)}\\
&= \lim_{h\to 0}\frac{2}{(x+2)(x+h+2)} \qquad\text{(since }h\neq 0\text{, so we can cancel)}\\
&= \frac{2}{(x+2)(x+0+2)} \qquad\text{(since the function is defined at }h=0\text{)}\\
&= \frac{2}{(x+2)^2}.
\end{align*}$$
