Find the leading order asymptotic behaviour of the integral $$I(x) = \int_0^{\infty}e^{-t-\frac{x}{t^2}}dt \mbox{ as } x \mbox{ tends to infinity} $$
I know this has a moveable maximum so you need to make a substitution which transforms it into the integral:
$$\frac{1}{x^{\frac{1}{3}}}\int_{0}^{\infty}\exp\left({-x^{\frac{1}{3}}\left(2^\frac{1}{3}s+\frac{1}{2^{\frac{2}{3}}s^2}\right)}\right)$$
but then what do I do from here? I don't know what substitution to make to reduce the integral further
 A: Your approach is on the right track.  But let's fix a few things.
In the original integral, let $t=2 ^{1/3}x^{1/3} s$; then the integral equals
$$2 ^{1/3} x^{1/3} \int_0^{\infty} ds \, \exp{\left [-x^{1/3} \left ( 2^{1/3} s + 2^{-2/3} s^{-2} \right )\right ]} $$
You do not need to evaluate this integral.  Laplace's method simply requires, at least for the leading behavior, finding the value(s) of $s$ that maximizes the argument of the exponential.  In this case, you can show that the max occurs at $s=1$.  In this case, we now rewrite the integral over a small region about $s=1$ (because the integral elsewhere is exponentially small).  Thus, let $s=1+u$ and expand out to $O(u^2)$; then we have the integral is approximately
$$2 ^{1/3} x^{1/3} \, e^{-3 \cdot 2^{1/3} x^{1/3}} \int_{-\epsilon}^{\epsilon} du \, e^{-3 \cdot 2^{-2/3} x^{1/3} u^2} $$
We can actually express bounds for $\epsilon$ in terms of the neglected, higher-order terms in the exponential.  However, because the integral is negligible outside of $[-\epsilon,\epsilon]$, we may replace the above integral with an integral over the real line and only introduce exponentially small error:
$$2 ^{1/3} x^{1/3} \, e^{-3 \cdot 2^{1/3} x^{1/3}} \int_{-\infty}^{\infty} du \, e^{-3 \cdot 2^{-2/3} x^{1/3} u^2} $$
This is a standard gaussian integral, which evaluates to a simple expression.  Thus, the leading asymptotic behavior of the integral is
$$2^{2/3} \sqrt{\frac{\pi}{3}} x^{1/6} \, e^{-3 \cdot 2^{-2/3} x^{1/3}} \quad (x \to \infty)$$
ADDENDUM
It is good practice to check the correctness of the approximation against numerical results.  A good check is a log-log plot of the relative error.  Here is an example:

The abscissa represents the log base $10$ of $x$, while the ordinate represents the log base $10$ of the absolute value of the relative error between a numerical evaluation of the integral and the derived asymptotic approximation.  The Mathematica code used to generate the plot is as follows:
ListLinePlot[Table[{x, Log[10, Abs[1 - 2^(2/3) Sqrt[Pi/3] (10^x)^(1/6) Exp[-3 2^(-2/3) (10^x)^(1/3)]/NIntegrate[Exp[-t - 10^x/t^2], {t, 0, Infinity}] ]]}, {x, 2, 4, 0.2}]]
Note that the slope of the resulting line is $-1/3$, which is exactly what we would expect in a log-log plot of the relative error, as the relative error indeed is $O(x^{-1/3})$.  (See Claude's CAS result, for example. If interested, I can illustrate how to derive higher-order results.)
A: Ron Gordon provided an elegant answer and, please, consider this more as a comment to long to fit than as an answer.
Using a CAS, I found that $$I(x)=\frac{G_{0,3}^{3,0}\left(\frac{x}{4}|
\begin{array}{c}
 0,\frac{1}{2},1
\end{array}
\right)}{\sqrt{\pi }}$$ in which appears  the Meijer $G$ function.
Expanding this result for large values of $x$, a CAS obtained $$I(x)= \sqrt{\frac{\pi }{3}} e^{-\frac{3 \sqrt[3]{x}}{2^{2/3}}} \left(2\ 2^{2/3} 
   \sqrt[6]{x}+\frac{5}{9} \sqrt[3]{2}
   \sqrt[6]{\frac{1}{x}}-\frac{35}{324} 
   \sqrt{\frac{1}{x}}+O\left(\left(\frac{1}{x}\right)^{13/6}\right)\right)$$
