Finding the derivative of a piecewise function

I am given a function f(x) = e^(-1/x) when x>0
0 when x <= 0.
The question tells me to compute f '(0).

I know I'm supposed to find a right hand derivative and a left hand derivative. The left is just 0, but I'm really struggling with the right side. Using the definition of a derivative, I started the problem as shown below:

lim x->0+ [f(x) - f(0)]/(x-0) = lim x->0+ [(e^(-1/x)-0)/(x-0)] = lim x->0+ e^(-1/x)/x^2

Then, I tried to evaluate the limit using L'Hopital's rule, but I keep getting indeterminate forms. Where am I messing up?

You have the right idea.
f '(0) = lim(x→0) [f(x) - f(0)] / (x - 0) = lim(x→0) f(x)/x.

Looking at one-sided limits:
lim(x→0-) f(x)/x = lim(x→0-) 0/x = 0.

lim(x→0+) f(x)/x
= lim(x→0+) e^(-1/x)/x
= lim(x→0+) (1/x) / e^(1/x), via double inversions
= lim(x→0+) (-1/x^2) / [(-1/x^2) e^(1/x)], by L'Hopital's Rule
= lim(x→0+) 1/e^(1/x)
= lim(x→0+) e^(-1/x)
= 0.

Hence, f '(0) exists and equals 0 (since the one-sided limits are equal).

I hope this helps!