Differentiation by first principles

How can I find the derivative of f(x)=e^(1/x^2) from first principles (ie limit as h->0)?

1) f(x) = e^(1/x²); ==> f(x+h) = e^{1/(x+h)²}

2) So, f(x+h) - f(x) = e^{1/(x+h)²} - e^(1/x²) = e^(1/x²)[e^{1/(x+h)² - 1/x²} - 1]

Further simplifications, = e^(1/x²)[e^(t) - 1], where t = {-h(2x+h)/x²(x+h)²}

3) dy/dx = Limit (h --> 0) [{f(x+h) - f(x)}/h]

= Limit (h --> 0) e^(1/x²)[e^(t) - 1]/h

= e^(1/x²)*[limi (h --> 0){e^t - 1}/h]

= e^(1/x²)*[Lim (h --> 0, t --> 0) {(e^t - 1)/t}*(t/h)]

= e^(1/x²)*[Lim (t --> 0) {(e^t - 1)/t}]* Lim (t,h --> 0)(t/h)

4) Applying Exponential limit theorem, Lim (t --> 0) {(e^t - 1)/t} = 1

Substituting for t, Lim (t, h --> 0) (t/h) = Lim(h-->0){-h(2x+h)/x²(x+h)²h}

= Lim (h --> 0) [-(2x+h)/{x²(x+h)²}] = -2/x³ [evaluating for the limit h = 0]

Thus substituting all these in (3) above,

dy/dx = (-2/x³)*{e^(1/x²)}

Use the definition of derivative as a limit ....

The rest is just algebra.