What does this function mean? 
$$f(x) = \frac{x}{e^{x^2}}$$
  Differentiate $f(x)$.

How should the above function be interpreted? Is the function equivalent to:

a)$$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{(e^x)^2}} = \frac{x}{e^{2x}}$$

or

b) 
  $$f(x) = \frac{x}{e^{x^2}} = \frac{x}{{e^{(x^2)}}} ≠ \frac{x}{e^{2x}}$$

Thanks!
 A: $a^{b^c}$ should be interpreted as $a^{(b^c)}$.  Because, as you point out, $(a^b)^c = a^{bc}$.  This is even true in most programming languages, which specify that a^b^c should be evaluated from right to left, unlike expressions like a-b+c, which should be evaluated from left to right.
A: The expression
$f(x) = x / e^{x^2} \tag{1}$
should be interpreted in the same way we would interpret $x / exp(x^2)$;that is,
$f(x) = x e^{-x^2} = x \exp (-x^2). \tag{2}$
The tip-off here is that the exponent $2$ of $x$ in the expression $x^2$ is in fact written as a superscript of $x$; in item (a) of the question, $(e^x)^2 = (e^x)(e^x) = e^{2x}$;  but
$e^{x^2}$ should be thought of as $e^{xx} = (e^x)^x$; as Stephen Montgomery points out in his answer, the evaluation of the exponents is right to left.
Having hopefully clarified the issue, we have
$f'(x) =  (x e^{-x^2})' = e^{-x^2} + x e^{-x^2} (-2x) \tag{3}$
using the Leibniz product rule and the formula $d / dx (e^u) = e^u (du / dx)$.  After the simple algebra,
$f'(x) = e^{-x^2}(1 - 2x^2). \tag{4}$
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
