A student forgot the Product Rule for differentiation and made the mistake of thinking that $(fg)'=f'g'$... 

A student forgot the Product Rule for differentiation and made the mistake of thinking
    that $(fg)'=f'g'$. However, he was lucky and got the correct answer. The function f that he used was $f(x)=e^{x^2}$
     and the domain of his problem was the interval $(\frac{1}{2} , \infty)$What was
    the function $g$?


For the product rule
$(fg)'=f'g + fg'=e^{x^2}\cdot 2x\cdot g+e^{x^2}\cdot g'$
What the student did
$(fg)'=f'g'=e^{x^2}\cdot 2x \cdot g'$
Now we assume both of these to be equivalent to other
$$(fg)' =(fg)'$$
$$ e^{x^2}\cdot 2x\cdot g+e^{x^2}\cdot g'=e^{x^2}\cdot 2x \cdot g'$$
$$  e^{x^2}\cdot 2x\cdot g = g' (e^{x^2} \cdot 2x - e^{x^2} )$$
$$ \frac{e^{x^2}\cdot 2x}{e^{x^2} \cdot 2x - e^{x^2}} = \frac{g'}{g} $$
$$ \frac{2x}{2x-1} = \frac{g'}{g} $$
This is the part I am unsure about since I'm not sure if I can cancel the $e^{x^2} out $ 
After dominiks comment I continue
$$ \frac{2x-1+1}{2x-1} = \frac{g'}{g} $$
$$ 1 + \frac{1}{2x-1} = \frac{g'}{g} $$
$$ \int 1 + \frac{1}{2x-1} dx = \int \frac{g'}{g} dx$$
$$ x+\frac{\ln|2x-1|}{2}+c =\ln|g|$$
$$ g=e^{x+\frac{\ln|2x-1|}{2}+c}$$
$$ \therefore g(x)={\sqrt{2x-1}} \cdot Ae^{x} $$

As stated in the comments, your cancellation of $e^{x^2}$ was fine.  However, you've made a mistake finding $g$.  You should find
$$
g=e^{x+\frac{\ln|2x-1|}{2}+c} = e^ce^x\left[e^{\ln|2x - 1|} \right]^{1/2} = Ae^{x} \sqrt{|2x-1|}
$$
and, since $x > 1/2$, the absolute value is redundant.
