How do I find the critical points of this function? My online homework question reads: 
Let $f(x) = (x - 3)^3 + (x - 3)^{-3}$. 

I set up to find the critical points by first finding the derivative of the equation. I used the power rule at first and got $3(x-3)^2-\frac{3}{(x-3)^4}$. Then I was attempting to use the power rule, however, I think I am going about the problem all wrong. 
Any assistance on how to find the critical points for the equation would be helpful! Thank you
You're on the right track.  Set the first derivative equal to zero:

$$0 = 3(x-2)^2 - \frac{3}{(x-3)^4}$$
$$(x-3)^6 = 1 \to x = 4, 2$$

Then continue from there.
the critical points $f'(x)=3(x-3)^2-\frac{3}{(x-3)^4}=0$

$3(x-3)^2=\frac{3}{(x-3)^4}$

$(x-3)^6=1$

$x=3$

Or graphically