Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$ 
Find all integer solutions $(x, y)$ of the equation 
  $$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$

What have done is that: 
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so, 
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we choose integer $y$ 
to make $2y-3=1 \text{ or } {-1}$. $y = 2$ or $y = 1$ is such a solution.
Also, $2y - 3$ can be deleted by numerator $3$, so $2y - 3$ can be $3$ or $-3$ too. This gives $y = 3$ or $y = 0$, but $y$ can not be $0$. So far, we have $y=1,2,3$. Finally, $2y-3$ can be deleted by numerator $y$, but how can we find such a $y$? 

Your approach can be continued to a full solution. Let's see it in detail:
Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:
  At least one of $x,y$ is positive, and by symmetry we can assume that $y$ is positive.
  $\frac{1}{x} = \frac{2y-3}{3y}$.
  Thus $x = \frac{3y}{2y-3} = 1 + \frac{y+3}{2y-3}$.
  If $y > 6$ then $y+3 < 2y-3$ and so $0 < \frac{y+3}{2y-3} < 1$, which makes $x$ not an integer.
  Thus $y \le 6$.
It only remains to check all $y$ from $1$ to $6$.
