Question: Is there any particular inequality involving $e^{x^2}$ that I need to use here? In any case, I couldn't solve it.

Answer: 
$$\int\_{0}^{x}e^{x^2-t^2}\,dt \stackrel{t=xs}{=} x\int\_{0}^{1}e^{x^2(1-s^2)}\,ds > x\int\_{0}^{1}1\,ds = x.$$
The rectangle $[0,x]\times[0,e^{-x^2}]$ lies entirely below the graph of $e^{-x^2}$.