#### Answer

$y=-0.08x+0.54$

#### Work Step by Step

Let's start off by finding the derivative of the original function $f(x)=\frac{x}{1+x^2}$
We're going to have to use the quotient rule, the quotient rule states that:
$\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$
So using that, we can now find the derivative of $f(x)$. Let's start by finding the derivative of the numerator and denominator:
Numerator: $x$
Derivative of Numerator: $1$
Denominator: $1+x^2$
Derivative of Denominator: $2x$
So now let's just plug that into the quotient rule:
$f'(x)=\frac{(1)(1+x^2)-(2x)(x)}{(1+x^2)^2}=\frac{1+x^2-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}$
Now, we need to find the slope of the tangent line at $(3,0.3)$. Let's plug in $x=3$:
$f'(3)=\frac{1-(3)^2}{(1+(3)^2)^2}=\frac{1-9}{(1+9)^2}=\frac{-8}{100}=-0.08$
So now we know that the slope $m=-0.08$ at the point $(3,0.3)$. We can use the point-slope form to create the equation of a line.
$y-y_1=m(x-x_1)$
Now we can plug in our information:
$y-0.3=-0.08(x-3)$
$y-0.3=-0.08x+0.24$
$y=-0.08x+0.54$
And that's your final answer. Now you can graph both the original function and the tangent line on the same graph to verify your answer.