Find the derivative of the function $f(x)=x/(x^2+1)$ at a point a

question is,, find functions at an arbitrary point.. i know i can just use that rule,, but the thing is i have to use an arbitrary point

$x$ is already arbitrary. It doesn't get any more arbitrary by using a different letter as a variable name. The whole point of a function is to associate one element of a set that you provide with another element (of a possibly different set).

And $x$ is any number of the real numbers (probably at least, you didn't specify what set $x$ comes from) without specifying which one. It's not $12$, not $-245234234.4564576$, it's any element of the set that the function "accepts". After all, this is why we use letters to denote some variable, because it's not one distinct number.

Your answer is correct.

Notice, $$f(x)=\frac{x}{x^2+1}$$

$$\implies f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}$$ $$=\lim_{h\to 0}\frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h((x+h)^2+1)(x^2+1)}$$ $$=\lim_{h\to 0}\frac{x^3+hx^2+x+h-x^3-2hx^2-h^2x-x}{h((x+h)^2+1)(x^2+1)}$$ $$=\lim_{h\to 0}\frac{h-hx^2}{h((x+h)^2+1)(x^2+1)}$$ $$=\lim_{h\to 0}\frac{1-x^2}{((x+h)^2+1)(x^2+1)}$$$$=\frac{1-x^2}{((x+0)^2+1)(x^2+1)}=\frac{1-x^2}{(x^2+1)^2}$$ Now, substituting arbitrary value $x=a$ $$f'(a)=\frac{1-a^2}{(a^2+1)^2}$$