Question on Showing points of discontinuities of a function are removable (or not) The question is as follows:

Given function: $F(x,y)=\frac{x + 2y}{sin(x+y) - cos(x-y)}$
Tasks: 
a/ Find points of discontinuities
b/ Decide if the points (of
  discontinuities) from part a are removable

Here is my work so far:
(1) For part a, I think the points of discontinuities should have form $(0, \frac{\pi}{4} + n\pi)$ or $(\frac{\pi}{4} + n\pi, 0)$ , since they make the denominator undefined.  For convenience of part b, I choose to specifically deal with the point $(0, \frac{\pi}{4})$
(2) Recall definition:
A point of discontinuity $x_0$ is removable if the limits of the function under certain path are equal to each other, as they are "close" to $x_0$.  In particular, if the function is 1 dimensional, we get the notion of "left" and "right" limits.  But here we talk about paths of any possible direction.  However, these limits are not equal to $f(x_0)$, which can be defined or undefined.
(3) 
I'm having trouble of "finding" such paths @_@
I come across with these two, by fix x-coordinate and vary y-coordinate:
$F(x, x^2 - \frac{\pi}{4})$
and $F(x, x^2 - x - \frac{\pi}{4})$
They both have limit to be $\frac{\pi}{2\sqrt(2)}$ as x approaches 0 (by my calculation)
But what I can say about these results?  I feel that discontinuities of $F(x,y)$ should be not removable, but I don't know if my thought is correct. 
Would someone please help me on this question?
Thank you in advance ^^

Observe that F(x,y) =(x+2y) /(2*Cos(y+pi/4)*Sin(x - pi/4)).just use formula for sin(a) - sin(b) , and cos(b) = sin (pi/2 -b)
Thus the set of discontinuities are (x, pi/4 +n*pi) and (pi/4 + n*pi, y) for any x,y real. ie. they are lines parellel to x and y axis ie. a grid. 
So we have to  look at point of intersection of line x+2y = 0 and the above grid.
For example at (pi/4 , -pi/8)  F(x,y) =(x+2y) /(2*Cos(y+pi/4)*Sin(x - pi/4)) . Cos(-pi/8 + pi/4) is non zero. So consider Lim as (x,y)->(pi/4 , -pi/8) (x+2y) /Sin (x-pi/4) . Make substitution x' = x-pi/4 and y' = y+pi/8. we get Lim (x',y')->(0,0) x'+2y'/Sin(x'). consider curve y'=0 we get limit as 1 . while curve x'+2y' = 0 we get limit as 0.
You can try this in general on the other points, we get a expression of form Lim (x',y')->(0,0) x'+2y'/{+ or -}Sin(x')  or x'+2y'/ {+ or -}Sin (y') As in both cases limit never exists it will not be a removable singularity at any point.
