How do i use the chain rule twice when differentiating?

Ok so for this question we'll use this example: y = (e^{-2x^2} + 2)^1/5
We start off by making the expression inside the brackets equal to u. In other words u = e^{-2x^2} + 2. This also means y = u^1/5
Using the chain rule, we need to find du/dx and dy/du. dy/du = 1/5u^-4/5. Now to find du/dx we must use the chain rule again as e^{-2x^2} has a power within a power.
So lets say z = e^{-2x^2} . You can use any letter but for this example i'll use z. And v = -2x². Using the chain rule we get: dz/dx = -4xe^{-2x^2}
Now we can go back to when we used the chain rule the first time to differentiate the whole expression. du/dx = -4xe^{-2x^2} and dy/du = 1/5u^-4/5. Multiplying these together and replacing u with e^{-2x^2} + 2 gives us our final answer. dy/dx = 1/5(e^{-2x^2} + 2)^-4/5-4xe^{-2x^2}
Simplifying gives us dy/dx = -4xe^{-2x^2}/5(e^{-2x^2} + 2)^4/5