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I'm working on it now by carefully listing all the possibilities.

Is there a fast way to do this? And what's the answer anyways?

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select two horizontal lines from the available $8$ horizonal lines
$\dbinom{8}{2}$ ways

select two vertical lines from the available 7 vertical lines
$\dbinom{7}{2}$ ways

Required count of rectangles
$\dbinom{8}{2} \times \dbinom{7}{2}$

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    Sorry it is $\dbinom{8}{2} \times \dbinom{7}{2}.$ There are 8 horizontal lines and 7 vertical lines... – 2017-01-26
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    thanks. you are right. i missed it. corrected now. – 2017-01-26
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    Of course if the answerer had cared to explain the reasoning behind his choices, he would have noticed his mistake. But it looks like he rushed to answer first. The explanation is that each rectangle is characterized by its perimeter. – 2017-01-26
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This is the way I solved this problem before I read Kiran's answer that exposed a faster method: The grid that encloses the $7 \times 6$ rectangles consists of $8 \times 7$ lines and $8 \cdot 7 = 56$ points. Two of such points define the diagonal of a rectangle and therefore exactely one rectangle. There are $56 \choose 2$ pairs of points. But if two points lie on the same horizontal line or on the same vertical line this is a degenerated rectangle that we won't count. On a horizontal line we can choose $8 \choose 2$ pairs of points and there are $7$ horizontal lines. On a vertical line we can choose $7 \choose 2$ pairs of points and there are $8$ vertical lines. We have to subtract them from our number. The remaining number counts each rectangle twice because a rectangle has two diagonale. So there are

$$\left ({56 \choose 2} - 7 \cdot {8 \choose 2} - 8 \cdot {7 \choose 2}\right)/2$$ rectangles.