Let $A = k[x^3,x^2y,y^3] \subset k [x,y] $. I want to find the integral closure of $A$ in its field of fractions. To do so, I first want to find the field of fractions $\mathrm{Frac}(A)$.
I think that $\mathrm{Frac}(A) \neq k(x,y)$, since if we want to check if $x \in \mathrm{Frac}(A)$ we would have $x = \frac {x^{\alpha}y^{\beta}} {x^{\alpha-1}y^{\beta}} $ with $x^{\alpha}y^{\beta} = (x^3)^{a_1}(x^2y)^{a_2}(y^3)^{a_3}$, therefore: $\alpha = 3a_1 + 2a_2$ and $\beta = a_2 + 3a_3$.
On the other hand, I have $x^{\alpha-1}y^{\beta} = (x^3)^{b_1}(x^2y)^{b_2}(y^3)^{b_3}$
therefore: $\alpha-1 = 3b_1 + 2b_2$ and $\beta = b_2 + 3b_3$.
This system has no integer solutions, therefore $x \notin \mathrm{Frac}(A) $.
Now continuing with this method doesn't seem to be very effective. How can I compute $\mathrm{Frac}(A)$ more efficiently?
I do not know any general algorithm to compute the fraction field of an integral domain effectively in the sense of providing a minimal set of generators over a given domain. So usually one proceeds by determining 'small' candidates and by carefully extending them if necessary. Let $K$ the the field of fractions of the given ring. Then it is clear that $\frac{x^2y}{x^3}=\frac{y}{x}\in K$. Hence $k(\frac{y}{x})\subseteq K$. Now $y^3\not\in k(\frac{y}{x})$, therefore we are forced to consider $k(\frac{y}{x},y^3)\subseteq K$. Now $x^3=y^3(\frac{x}{y})^3\in k(\frac{y}{x},y^3)$ and $x^2y=(\frac{x}{y})^2y^3\in k(\frac{y}{x},y^3)$. Hence $K=k(x^3,x^2y,y^3)=k(\frac{y}{x},y^3)$.