if xy>0 and x>y prove that 1/y>1/x

This is how I would approach this -

I. Algebra:

First analyse x*y > 0. I take a mental note that this means that both x and y have the same sign.

I now move on to the second statement: x > y

I observe that if I divide both sides by x*y, I will have a format very similar to what is needed to be proven. I also see that x*y is always positive (it is after all given), so I do not need to reverse the signs.

So I divide both sides of the inequality by x*y, and get 1/y > 1/x.

II. Logic/Number Line:

Again, from the first statement I know that both x & y have the same sign.

I now visualise both x & y on the number line - And in both cases - whether both are positive or both are negative, their reciprocals will:

a. Remain on the same side of zero;
b. Will flip in relative distance from zero;

This will happen even if each of x or y are greater than 1 (or less than -1) or less than 1 (or greater than -1) - as long as both x & y are on the same side of zero, and not equal.

If you cannot 'visualise' this on the number line, then try testing cases. But eventually you ought to internalise the number line so that this mental visualisation becomes easier. Even if you have to test cases - doing so with a clear mental image of the number line will make the process that much less prone to errors and oversight.