I think that you mean that you want to use the definition of the derivative to find the derivative of the function $f(x)=\frac{x}{x+2}\;.$ The definition is that $f\,'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h\;.$ Note the limit: it’s essential.
For your function this is
$\begin{align*} \lim_{h\to 0}\frac{\frac{x+h}{(x+h)+2}-\frac{x}{x+2}}h&=\lim_{x\to 0}\left(\frac1h\left(\frac{x+h}{(x+h)+2}-\frac{x}{x+2}\right)\right)\\ &=\lim_{h\to 0}\left(\frac1h\cdot\frac{(x+2)(x+h)-x(x+2+h)}{(x+2)(x+2+h)}\right)\\ &=\lim_{h\to 0}\frac{x^2+2x+hx+2h-x^2-2x-hx}{h(x+2)(x+2+h)}\\ &=\lim_{h\to 0}\frac{2h}{h(x+2)(x+2+h)}\\ &=\lim_{h\to 0}\frac2{(x+2)(x+2+h)}\;, \end{align*}$
where I’ve simply put the original numerator $f(x+h)-f(x)$ over a common denominator and simplified. At this point we’ve managed to get rid of the factor of $h$ in the denominator, so we can take the limit: as $h\to 0$, the numerator just sits there at $2$, and the denominator approaches $(x+2)(x+2)$, or $(x+2)^2$. Thus,
$f\,'(x)=\lim_{x\to 0}\frac2{(x+2)(x+2+h)}=\frac2{(x+2)^2}\;.$
Added: By the way, if your algebra is good enough, you can notice that $\frac{x}{x+2}=\frac{(x+2)-2}{x+2}=1-\frac2{x+2}\;.$ Then
$\begin{align*} f(x+h)-f(x)&=\left(1-\frac2{(x+h)+2}\right)-\left(1-\frac2{x+2}\right)\\ &=\frac2{x+2}-\frac2{x+2+h}\\ &=\frac{2\big(x+2+h-(x+2)\big)}{(x+2)(x+2+h)}\\ &=\frac{2h}{(x+2)(x+2+h)}\;, \end{align*}$
and you reach the point of being able to take the limit a bit more easily.