Using L'Hopital's rule to find the limit of this function

I need to find the limit of this function using L’Hopitals rule and I am not sure about my calculations
$$ lim_{xto (-infty)}frac{e^frac{1}{x^3}-1}{arctanfrac{2}{x^4}}$$
Using Lhopital rule and finding the derivative I get
$$ lim_{xto (-infty)}frac{e^frac{1}{x^3}*frac{-3}{x^4} }{frac{1}{1+frac{4}{x^8}}* frac{-8}{x^5}}$$
but Im not sure how to proceed further I tried simplifying but Im not 100% confident that’s correct
$$ lim_{xto (-infty)}frac{e^frac{1}{x^3}*frac{-3}{x^4} }{frac{1}{1+frac{4}{x^8}}* frac{-8}{x^5}} = frac{e^frac{1}{x^3}*frac{-3}{1} }{frac{1}{1+frac{4}{x^8}}* frac{-8}{x}} = frac{e^frac{1}{x^3}*-3 }{frac{-8}{(1+frac{4}{x^8})x}} = frac{e^frac{1}{x^3}*-3 }{frac{-8}{x+frac{4}{x^7}}} = e^frac{1}{x^3}*-3 *{frac{x+frac{4}{x^7}}{-8}} = 1*-3 {frac{-infty+0}{-8}} = -infty$$