#### Answer

$f'(x) =1-9x^{-2}$
$slope = 0$
Equation of tangent line: $y = -6$

#### Work Step by Step

$f(x) = x +\frac{9}{x}$
$f(-3) = -3 -\frac{9}{3} = -6$
Now we know the point on the graph is: (-3, -6).
So,
$f'(x) = 1+9(-1)x^{-1-1}$
$f'(x) =1-9x^{-2}$
Therefore, the slope at x = -3 is:
$slope = 0$
So, the equation of tangent is:
$\frac{y+6}{x+3} = 0$
y+6 = 0
y = -6